Nice question! Assume that $p^2 \mid S$. This would then give $p \mid \varphi(S)$. However, this is impossible in our case since:
$$p \mid S^{S+\sigma(S)}-\varphi(S)\sigma(S^{\varphi(S+\sigma(S))}) \implies p \mid 1$$
which is clearly false.
Since $S$ is square-free, let:
$$S=\prod_{i=1}^n p_i$$
for distinct primes $p_1,p_2, \ldots ,p_n$. Then, by substitution:
$$1+\varphi(\prod p_i)\sigma(\prod p_i^{\varphi(S+\sigma(S))})=\prod p_i^{S+\sigma(S)}$$
$$1+\bigg(\prod (p_i-1)\bigg)\bigg(\prod \frac{p_i^{\varphi(S+\sigma(S))+1}-1}{p_i-1}\bigg)=\prod p_i^{S+\sigma(S)}$$
$$\bigg(\prod (p_i^{\varphi(S+\sigma(S))+1}-1)\bigg)=\bigg(\prod p_i^{S+\sigma(S)}\bigg)-1$$
But, we have:
$$\bigg(\prod (p_i^{\varphi(S+\sigma(S))+1}-1)\bigg) \leqslant \bigg(\prod (p_i^{S+\sigma(S)}-1)\bigg) \leqslant \bigg(\prod p_i^{S+\sigma(S)}\bigg)-1$$
We need equality in both places. The first equality can only hold true when $S+\sigma(S)$ is prime, so that $\varphi(S+\sigma(S))+1=S+\sigma(S)$. The second inequality can hold true only when there is one term in the product, i.e. $n=1$.
Since $N=1$, $S$ is prime. This shows that $S+\sigma(S)=2S+1$ is prime. Hence, $S$ is a Sophie-Germain prime, proving your conjecture to be true!
The two conjectures are true.
Conjecture 1 is true.
Proof :
$(1)$ is equivalent to
$$y^{\sigma(x)-x+1}=\varphi(x)$$
Suppose that $x$ is a composite number. Then, there exists a divisor $d$ of $x$ such that $\sqrt x\le d\lt x$, so we get $\sigma(x)\ge 1+\sqrt x+x$. So, we have
$$\varphi(x)=y^{\sigma(x)-x+1}\ge 2^{\sqrt x+2}$$
Here, let us prove that $2^{\sqrt x+2}\gt x$ for $x\gt 1$.
Let $f(x)=2^{\sqrt x+2}-x$. Then, we have $f'(x)=\frac{g(x)}{\sqrt x}$ where $g(x)=2^{\sqrt x+1}\ln 2-\sqrt x$. We have $g'(x)=\frac{h(x)}{2\sqrt x}$ where $h(x)=2^{\sqrt x+1}(\ln 2)^2-1$. Since $h(x)$ is increasing with $h(1)=\ln(4e)\ln(\frac 4e)\gt 0$, we get $h(x)\gt 0$ from which $g'(x)\gt 0$ follows with $g(1)=\ln\frac{16}{e}\gt 0$. Since $g(x)\gt 0$, we see that $f'(x)\gt 0$ with $f(1)=7\gt 0$ from which $f(x)\gt 0$ follows.$\quad\square$
So, we get
$$\varphi(x)=y^{\sigma(x)-x+1}\ge 2^{\sqrt x+2}\gt x\gt \varphi(x)$$
which is impossible.
So, $x$ has to be a prime number, and we get $y^{2}=x-1$.
It follows that $x$ has to be a near-square prime.$\quad\blacksquare$
Conjecture 2 is true.
Proof :
$(2)$ is equivalent to
$$(\sigma(x)-x+1)^y=\log_2\varphi(x)$$
The LHS is a positive integer, so there has to be a positive integer $k$ such that $\varphi(x)=2^k$. So, we see that $x$ has to be either of the form $x=2^m$ or of the form
$$x=2^m\prod_{i=1}^{n}(2^{a_i}+1)$$
where $2^{a_1}+1,2^{a_2}+1,\cdots, 2^{a_n}+1$ are distinct primes.
In the former, we get
$$(2^{m})^y=m-1$$
which is impossible since the LHS is larger than the RHS.
In the latter, suppose that $m\ge 1$. Then, we have
$$\begin{align}m-1+a_1+a_2+\cdots +a_n&=\bigg(1+(2^{m+1}-1)\prod_{i=1}^{n}(2^{a_i}+2)-2^m\prod_{i=1}^{n}(2^{a_i}+1)\bigg)^y
\\\\&\gt (2^{m+1}-1)\prod_{i=1}^{n}(2^{a_i}+2)-2^m\prod_{i=1}^{n}(2^{a_i}+1)
\\\\&\gt (2^{m+1}-1)\prod_{i=1}^{n}(2^{a_i}+1)-2^m\prod_{i=1}^{n}(2^{a_i}+1)
\\\\&= (2^m-1)\prod_{i=1}^{n}(2^{a_i}+1)\end{align}$$
from which we have
$$m-1+a_1+a_2+\cdots +a_n\gt (2^m-1)\prod_{i=1}^{n}(2^{a_i}+1)$$
which is impossible. So, we have to have $m=0$.
Suppose that $n\ge 2$. Then, we have
$$\begin{align}a_1+a_2+\cdots+a_n&=\bigg(1+\prod_{i=1}^{n}(2^{a_i}+2)-\prod_{i=1}^{n}(2^{a_i}+1)\bigg)^y
\\\\&\gt \prod_{i=1}^{n}(2^{a_i}+1+1)-\prod_{i=1}^{n}(2^{a_i}+1)
\\\\&\gt \sum_{i=1}^{n}(2^{a_i}+1)\end{align}$$
from which we have
$$a_1+a_2+\cdots+a_n\gt \sum_{i=1}^{n}(2^{a_i}+1)$$
which is impossible. So, we have to have $n=1$.
So, we have to have $x=2^{a_1}+1$ and $2^y=a_1$.
It follows that $x$ has to be a Fermat prime greater than $3$.$\quad\blacksquare$
Best Answer
The conjecture is true.
Proof :
$x=1$ does not satisfy $(3)$, and $x=2$ satisfies $(3)$. In the following, $x\ge 3$.
As you already noticed, $x$ has to be a square-free integer.
Then, letting $x=\displaystyle\prod_{k=1}^{n}p_k$ with $n=\omega(x)$, we get
$$\prod_{k=1}^{n}\bigg({p_k}^{\sigma(2^{1+\varphi(x)}-1)}-1\bigg)=-1+\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}$$
Suppose here that $2^{1+\varphi(x)}-1$ is a composite number.
Here, using the following facts :
If $N$ is a composite number, then $\sigma(N)\ge 1+\sqrt N+N$.
If $N\ge 3$, then $\varphi(N)\ge 2$.
If $m\ge 2$ and $y\gt 0$, then $m^{2+y}-1\ge m^{1+y}$.
we get
$$\begin{align}-1&=\prod_{k=1}^{n}\bigg({p_k}^{\sigma(2^{1+\varphi(x)}-1)}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&\ge \prod_{k=1}^{n}\bigg({p_k}^{1+\sqrt{2^{1+\varphi(x)}-1}+2^{1+\varphi(x)}-1}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&=\prod_{k=1}^{n}\bigg({p_k}^{\sqrt{2^{1+\varphi(x)}-1}+2^{1+\varphi(x)}}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&\ge\prod_{k=1}^{n}\bigg({p_k}^{\sqrt{2^{1+2}-1}+2^{1+\varphi(x)}}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&\ge\prod_{k=1}^{n}\bigg({p_k}^{2+2^{1+\varphi(x)}}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&\ge\prod_{k=1}^{n}\bigg({p_k}^{1+2^{1+\varphi(x)}}\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}\end{align}$$ from which we have $$-1\ge \prod_{k=1}^{n}\bigg({p_k}^{1+2^{1+\varphi(x)}}\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}$$ which is impossible since the RHS is positive.
So, we see that $2^{1+\varphi(x)}-1$ is a prime number.
Then, we get $$\prod_{k=1}^{n}\bigg({p_k}^{2^{1+\varphi(x)}}-1\bigg)=-1+\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}$$
Suppose here that $n\ge 2$. Then, we get $$\begin{align}1&=\prod_{k=1}^{n}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)-\prod_{k=1}^{n}\bigg({p_k}^{2^{1+\varphi(x)}}-1\bigg) \\\\&\ge {p_n}^{2^{1+\varphi(x)}}\prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)-\bigg({p_n}^{2^{1+\varphi(x)}}-1\bigg)\prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg) \\\\&=\prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)\end{align}$$ from which we have $$1\ge \prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)$$ which is impossible since the RHS is larger than $1$.
So, we have to have $\omega(x)=n=1$, and $x$ has to be a prime number.
Therefore, $x$ has to be a Mersenne exponent. $\quad\blacksquare$