If a positive integer $n$ satisfies
$$4\psi(n)^2-(12n+3)\psi(n)+9n^2=0\tag1$$ then $n$ is an even perfect number.
Proof :
Solving $(1)$ for $\psi(n)$ gives
$$\psi(n)=\frac{12n+3\pm 3\sqrt{1+8n}}{8}$$
There has to be a positive integer $k$ such that $1+8n=k^2$, so
$$\psi\bigg(\frac{k^2-1}{8}\bigg)=\frac{12\cdot\frac{k^2-1}{8}+3\pm 3k}{8}=\frac 3{16}(k\pm 1)^2$$
Since $k$ has to be odd, there has to be a positive integer $m$ such that $k=2m+1$ to have
$$\psi\bigg(\frac{m(m+1)}{2}\bigg)=\frac 3{4}\bigg(m+\frac{1\pm 1}{2}\bigg)^2$$
There has to be a positive integer $\ell$ such that $m+\dfrac{1\pm 1}{2}=2\ell$, so
$$\psi\bigg(\ell(2\ell\mp 1)\bigg)=3\ell^2$$
Since $\psi$ is the multiplicative function with $\gcd(\ell,2\ell\mp 1)=1$,
$$\psi(\ell)\psi(2\ell\mp 1)=3\ell^2$$
For $\ell=1$, this does not hold. Suppose here that $\ell\gt 1$ is odd. Then, LHS is even while RHS is odd, which is impossible. So, $\ell$ has to be even, and there has to be a positive integer $a,s$ ($s$ is odd) such that $\ell=2^as$, so
$$\psi(2^a)\psi(s)\psi(2^{a+1}s\mp 1))=3\cdot 2^{2a}s^2,$$
i.e.
$$3\cdot 2^{a-1}\psi(s)\psi(2^{a+1}s\mp 1)=3\cdot 2^{2a}s^2,$$
i.e.
$$\psi(s)\psi(2^{a+1}s\mp 1)=2^{a+1}s^2$$
Suppose here that there are $a,s$ such that $\psi(s)\psi(2^{a+1}s+1)=2^{a+1}s^2$. Then, LHS is larger than RHS, which is impossible.
So, we have to have
$$\psi(s)\psi(2^{a+1}s-1)=2^{a+1}s^2$$
Let $\displaystyle 2^{a+1}s-1=\prod_{i=1}^{N} p_i^{b_i}$ where $p_i$ are distinct primes and $b_i$ are positive integers. Then, $$\psi(s)\prod_{i=1}^{N} p_i^{b_i-1}(p_i+1)=\bigg(1+\prod_{i=1}^{N} p_i^{b_i}\bigg)s$$Since RHS is not divisible by any $p_i$, we have to have $b_i=1$ for which we have $$\psi(s)\prod_{i=1}^{N} (p_i+1)=\bigg(1+\prod_{i=1}^{N} p_i\bigg)s$$Suppose here that $N\geqslant 2$. Then, LHS is larger than RHS, which is impossible. Therefore, $N=1$ implies $\psi(s)=s$, i.e. $s=1$. So, $n$ has to be of the form $n=2^a(2^{a+1}-1)$ where $2^{a+1}-1$ is prime. This means that $n$ has to be an even perfect number. $\quad\blacksquare$
This is a partial answer.
This answer proves the following claims :
Claim 1 : If $y=2x$ with odd $x$, then $(1)$ does not hold.
Claim 2 : If $y\not=2x$ and $y\not=2x-1$, then $y$ is a composite number.
Claim 3 : If $y\lt 3x-1$ and $y\not=2x$, then $x$ is a composite number.
Let us use the following lemma :
Lemma : If $n\ge 2$, then $P(\psi(n))\le n+1$. If $n=3$ or $n\ge 5$, then $P(\psi(n))\le \frac{n+1}{2}$ where $P(n)$ is the largest prime factor of $n$.
Proof for lemma : If $n=2,4$, then $P(\psi(n))=3\le n+1$.
If $n=2^k$ with $k\ge 3$, then
$$P(\psi(n))=P(\psi(2^k))=P(3\cdot 2^{k-1})=3\le\frac{n+1}{2}$$
If $n=p$ where $p$ is an odd prime, then
$$P(\psi(n))=P(\psi(p))=P(p+1)\le \frac{p+1}{2}=\frac{n+1}{2}$$
If $n=p^k$ where $p$ is an odd prime with $k\ge 2$, then
$$P(\psi(n))=P(\psi(p^k))=P(p^{k-1}(p+1))=p\le\frac n3\le\frac{n+1}{2}$$
If $\displaystyle n=\prod_{i=1}^{d}p_i^{e_i}$ where $d\ge 2$ and $p_1\lt p_2\lt \cdots\lt p_d$ are primes, then
$$P(\psi(n))=P\bigg(\psi\bigg(\prod_{i=1}^{d}p_i^{e_i}\bigg)\bigg)=P\bigg(\prod_{i=1}^{d}p_i^{e_i-1}(p_i+1)\bigg)\le p_d\le\frac n2\le\frac{n+1}{2}$$ since $p_d+1$ is an even composite number. $\quad\blacksquare$
Claim 1 : If $y=2x$ with odd $x$, then $(1)$ does not hold.
Proof : $(x,y)=(1,2)$ is not a solution. If $x$ is odd larger than $1$, then we get
$2\psi(x)=3\psi(x)$ which does not hold. $\quad\blacksquare$
Claim 2 : If $y\not=2x$ and $y\not=2x-1$, then $y$ is a composite number.
Proof : Since $\psi(x^y y^x)=x^{y-1}y^{x-1}\psi(xy)$, the equation $(1)$ is equivalent to
$$y^{y-x}(y-x)^{y-1}\psi(y-x)=x^{y-1}y^{x-1}\psi(xy)$$
Suppose that $y$ is a prime number.
Case 1 : $y\le 2x-2$. Then, we have$$(y-x)^{y-1}\psi(y-x)=x^{y-1}y^{2x-y-1}\psi(xy)$$So, $\psi(y-x)$ has to be divisible by $y$. If $y-x=1$, then since $x\ge 3$, we see that $1=x^{x}(x+1)^{x-2}\psi(x(x+1))$ has no solutions. It follows from $y-x\ge 2$ and Lemma that $P(\psi(y-x))\le y-x+1$. If $x\ge 2$, then $P(\psi(y-x))\le y-1$ from which $y\not\mid \psi(y-x)$ follows. So, we get $x=1$ and $y\le 0$ which is impossible.
Case 2 : $y\gt 2x$. Then, we have
$$y^{y-2x+1}(y-x)^{y-1}\psi(y-x)=x^{y-1}\psi(x)(y+1)$$
So, $\psi(x)$ has to be divisible by $y$. If $x=1$, then $y^{y-1}(y-1)^{y-1}\psi(y-1)=y+1$ which has no solutions since for $y\ge 3$, then LHS is larger than RHS. It follows from $x\ge 2$ and Lemma that $P(\psi(x))\le x+1$. If $y-x\ge 2$, then $P(\psi(x))\le x+1\le y-1$ from which $y\not\mid\psi(x)$ follows. So, we have to have $y-x=1$ and $x\lt 1$ which is impossible.
From the two cases above, the claim follows. $\quad\blacksquare$
Claim 3 : If $y\lt 3x-1$ and $y\not=2x$, then $x$ is a composite number.
Proof : Suppose that $x$ is a prime number.
Case 1 : $y\le 2x-2$. Then, we have$$(y-x)^{y-1}\psi(y-x)=x^{y-1}y^{2x-y-1}\psi(xy)$$
If $y-x=1$, then $1=x^{x}(x+1)^{x-2}\psi(x)\psi(x+1)$ which is impossible. Since $y-x\ge 2$, we have $y-x\lt x$ and $P(\psi(y-x))\le y-x+1\lt x$, and so LHS is not divisible by $x$.
Case 2 : If $y=2x-1$, then
$$(x-1)^{2x-2}\psi(x-1)=x^{2x-2}\psi(x)\psi(2x-1)$$
So, $\psi(x-1)$ has to be divisible by $x$. From Lemma, if $x-1=3$ or $x-1\ge 5$, then $P(\psi(x-1))\le\frac{x}{2}\lt x$, so we have to have $x-1=1,2,4$ for which the equation does not hold.
Case 3 : $2x\lt y\lt 3x-1$. Then, we have
$$y^{y-2x+1}(y-x)^{y-1}\psi(y-x)=x^{y-1}\psi(x)(y+1)$$
If $y-x\not=1,2,4$, then $P(\psi(y-x))\le\frac{y-x+1}{2}\lt x$. So, since $x$ is prime, we have to have $y=kx$ to have $2\lt k\lt 3-\frac 1x$ which is impossible. If $y-x=1$, then $(x+1)^{2-x}=x^{x}\psi(x)(x+2)$ has no solutions $x\le 2$. If $y-x=2$, then $(x+2)^{3-x}\cdot 2^{x+1}\cdot 3=x^{x+1}\psi(x)(x+3)$ has no solutions $x\le 3$. If $y-x=4$, then $(x+4)^{5-x}\cdot 2^{2x+6}\cdot 6=x^{x+3}\psi(x)(x+5)$ has no solutions $x\le 5$.
From the three cases above, the claim follows. $\quad\blacksquare$
Best Answer
Added : I've just added a proof of another claim about an upper bound for $N_k$ at the end.
This answer proves the following claim :
Claim : $$ N_k=\begin{cases}2 &\text{$\quad $if $\ k=2$} \\\\4&\text{$\quad$if $\ k=3$}\\\\ 16&\text{$\quad$if $\ k=4$} \end{cases}$$
The only solutions are $$\begin{align}k=2& : n=1,2^13^2 \\\\ k=3 & : n=1,2^{2}3^{3},2^13^25^{4},2^13^117^{4} \\\\ k=4 & : n=1,2^{3}3^{4},2^{2}3^{3}5^5,2^{2}3^{2}17^5,2^{2}3^{1}53^5, \\&\qquad\quad 2^13^311^{5},2^{1}3^{1}107^5,2^13^15^{5}17^{5},2^13^25^{4}29^{5}, \\&\qquad\quad 2^13^15^{4}89^{5},2^13^{2}5^{3}149^{5},2^13^{1}5^{3}449^{5},2^13^{3}5^{1}1249^{5}, \\&\qquad\quad 2^13^{1}17^{4}101^{5},2^13^{2}17^{3}577^{5},2^13^{1}17^{3}1733^{5}\end{align}$$
Proof :
$n=1$ is a solution for $(2)$.
For an odd prime $p$, the numerator of $\frac{p+1}{p}$ is even. This implies that if $n$ is odd larger than $1$, then the equation $(2)$ does not hold. So, $n$ has to be even, and then $n$ has a prime factor $3$.
If $n=2^s3^t$ where $s,t\ge 1$, then $(2)\implies 2^{s+1}3^t=2^k3^k\implies n=2^{k-1}3^{k}$.
If $n=2^s3^t\prod_{j=1}^{d}p_j^{e_j}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are odd primes larger than $3$, and $d,s,t,e_j$ are positive integers. Then, $(2)$ is equivalent to
$$2^s3^t\bigg(\prod_{j=1}^{d}p_j^{e_j}\bigg)\cdot\frac 32\cdot\frac 43\prod_{j=1}^{d}\bigg(1+\frac{1}{p_j}\bigg)=2^k3^k\prod_{j=1}^{d}p_j^k$$ which can be written as $$\prod_{j=1}^{d}(p_j+1)=2^{k-1-s}3^{k-t}\prod_{j=1}^{d}p_j^{k+1-e_j}$$ where we have to have $s\le k-1, t\le k$ and $e_j\le k+1$.
Since LHS is divisible at least by $2^d$, we have to have $1\le d\le k-1-s\le k-2$ implying $k\ge 3$.
$k=2$ :
The only solutions are $n=1,2^13^2$, and so $N_2=2$.
$k=3$ :
$n=1,2^{2}3^{3}$ are solutions.
If $n=2^s3^t\prod_{j=1}^{d}p_j^{e_j}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are odd primes larger than $3$, and $d,s,t,e_j$ are positive integers. Then, the equation is equivalent to
$$\prod_{j=1}^{d}(p_j+1)=2^{2-s}3^{3-t}\prod_{j=1}^{d}p_j^{4-e_j}$$ where we have to have $s\le 2, t\le 3$ and $e_j\le 4$.
Since LHS is divisible at least by $2^d$, we have to have $1\le d\le 2-s\le 1$ implying $d=1$ for which we have
$$p_1+1=2^{2-s}3^{3-t}p_1^{4-e_1}$$ Since $4-e_1=0$, we get $p_1=2^{2-s}3^{3-t}-1$ with $s=1$.
$2^{1}3^{1}-1=5$ is a prime, and $n=2^13^25^{4}$.
$2^{1}3^{2}-1=17$ is a prime, and $n=2^13^117^{4}$.
Therefore, it follows that $N_3=4$.
$k=4$ :
$n=1,2^{3}3^{4}$ are solutions.
If $n=2^s3^t\prod_{j=1}^{d}p_j^{e_j}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are odd primes larger than $3$, and $d,s,t,e_j$ are positive integers. Then, the equation is equivalent to
$$\prod_{j=1}^{d}(p_j+1)=2^{3-s}3^{4-t}\prod_{j=1}^{d}p_j^{5-e_j}$$ where we have to have $s\le 3, t\le 4$ and $e_j\le 5$.
Since LHS is divisible at least by $2^d$, we have to have $1\le d\le 3-s\le 2$.
Case 1 : $d=1$
$$p_1+1=2^{3-s}3^{4-t}p_1^{5-e_1}$$
Since $5-e_1=0$, we have $p_1=2^{3-s}3^{4-t}-1$.
$2^{1}3^{1}-1=5$ is a prime, and $n=2^{2}3^{3}5^5$.
$2^{1}3^{2}-1=17$ is a prime, and $n=2^{2}3^{2}17^5$.
$2^{1}3^{3}-1=53$ is a prime, and $n=2^{2}3^{1}53^5$.
$2^{1}3^{4}-1=161$ is not a prime.
$2^{2}3^{1}-1=11$ is a prime, and $n=2^13^311^{5}$.
$2^{2}3^{2}-1=35$ is not a prime.
$2^{2}3^{3}-1=107$ is a prime, and $n=2^{1}3^{1}107^5$
$2^{2}3^{4}-1=323$ is not a prime.
Case 2 : $d=2$
Since $s=1$, we have $$(p_1+1)(p_2+1)=2^{2}3^{4-t}p_1^{5-e_1}p_2^{5-e_2}$$ Now, $5-e_2=0$, and there is a non-negative integer $a$ such that $$p_1+1=2^13^{a}\qquad\text{and}\qquad p_2+1=2^13^{4-t-a}p_1^{5-e_1}$$
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{0}-1=17$ is a prime, and $n=2^13^15^{5}17^{5}$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{0}5^{1}-1=9$ is not a prime.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{1}5^{1}-1=29$ is a prime, and $n=2^13^25^{4}29^{5}$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{1}-1=89$ is a prime, and $n=2^13^15^{4}89^{5}$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^13^{0}5^{2}-1=49$ is not a prime.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{1}5^{2}-1=149$ is a prime, and $n=2^13^{2}5^{3}149^{5}$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{2}-1=449$ is a prime, and $n=2^13^{1}5^{3}449^{5}$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{0}5^{3}-1$ is not a prime with $3\mid p_2$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{1}5^{3}-1$ is not a prime with $7\mid p_2$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{3}-1$ is not a prime with $13\mid p_2$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{0}5^{4}-1=1249$ is a prime, and $n=2^13^{3}5^{1}1249^{5}$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{1}5^{4}-1$ is not a prime with $23\mid p_2$.
$p_1=2^13^{1}-1=5$ is a prime and $p_2=2^1 3^{2}5^{4}-1$ is not a prime with $7\mid p_2$.
$p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{0}17^{1}-1=33$ is not a prime.
$p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{1}17^{1}-1=101$ is a prime, and $n=2^13^{1}17^{4}101^{5}$.
$p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{0}17^{2}-1=577$ is a prime, and $n=2^13^{2}17^{3}577^{5}$.
$p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{1}17^{2}-1=1733$ is a prime, and $n=2^13^{1}17^{3}1733^{5}$.
$p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{0}17^{3}-1$ is not a prime with $5\mid p_2$.
$p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{1}17^{3}-1$ is not a prime with $7\mid p_2$.
$p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{0}17^{4}-1$ is not a prime with $7\mid p_2$.
$p_1=2^13^{2}-1=17$ is a prime and $p_2=2^1 3^{1}17^{4}-1$ is not a prime with $5\mid p_2$.
$p_1=2^13^{3}-1=53$ is a prime and $p_2=2^1 3^{0}53^{1}-1$ is not a prime with $5\mid p_2$.
$p_1=2^13^{3}-1=53$ is a prime and $p_2=2^1 3^{0}53^{2}-1$ is not a prime with $41\mid p_2$.
$p_1=2^13^{3}-1=53$ is a prime and $p_2=2^1 3^{0}53^{3}-1$ is not a prime with $3\mid p_2$.
$p_1=2^13^{3}-1=53$ is a prime and $p_2=2^13^{0}53^{4}-1$ is not a prime with $7\mid p_2$.
Therefore, it follows that $N_4=16$.
I'm going to prove the following claim about an upper bound for $N_k$.
Claim 2 : For $k\ge 5$, $$N_k\le 2+\sum_{d=1}^{k-2}(k-2)^d(k-1)k^{d+1}(k+1)^{\frac{d(d+1)}{2}}$$
Proof :
We already know that $n=1,n=2^{k-1}3^{k}$ are solutions.
If $n=2^s3^t\prod_{j=1}^{d}p_j^{e_j}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are odd primes larger than $3$, and $d,s,t,e_j$ are positive integers. Then, $(2)$ is equivalent to
$$\prod_{j=1}^{d}(p_j+1)=2^{k-1-s}3^{k-t}\prod_{j=1}^{d}p_j^{k+1-e_j}$$ where we have to have $s\le k-1, t\le k$ and $e_j\le k+1$.
Since LHS is divisible at least by $2^d$, we have to have $1\le d\le k-1-s\le k-2$ implying $k\ge 3$.
We can write $$\begin{cases}p_1+1&=2^{a_1}3^{b_1} \\\\ p_2+1&=2^{a_2}3^{b_2}p_1^{c(2,1)} \\\\ p_3+1&=2^{a_3}3^{b_3}p_1^{c(3,1)}p_2^{c(3,2)} \\\\\qquad\vdots \\\\p_d+1&=2^{a_d}3^{b_d}p_1^{c(d,1)}p_2^{c(d,2)}\cdots p_{d-1}^{c(d,d-1)}\end{cases}$$ where $1\le a_j\le k-2,0\le b_j\le k-1$ and $0\le c(j,i)\le k$.
The number of possible $p_1$ is at most $(k-2)k$.
For each $p_1$, the number of possible $p_2$ is at most $(k-2)k(k+1)$.
For each pair $(p_1,p_2)$, the number of possible $p_3$ is at most $(k-2)k(k+1)^2$.
So, we see that the number of possible $(p_1,p_2,\cdots,p_d)$ is at most $$\prod_{j=1}^{d}(k-2)k(k+1)^{j-1}$$
For each $(p_1,p_2,\cdots,p_d)$, the number of possible $n$ is at most $$(k-1)k(k+1)^d$$
Therefore, we get, for $k\ge 5$, $$\begin{align}N_k&\le 2+\sum_{d=1}^{k-2}(k-1)k(k+1)^d\prod_{j=1}^{d}(k-2)k(k+1)^{j-1} \\\\&=2+\sum_{d=1}^{k-2}(k-2)^d(k-1)k^{d+1}(k+1)^{\frac{d(d+1)}{2}}\end{align}$$