We give an "examination of all cases" solution. Use the fact that $\varphi(n)$ is multiplicative. Let
$$n=2^a p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k},$$
where the $p_i$ are distinct odd primes, the $e_i$ are $\ge 1$, and $a \ge 0$.
Then
$$\varphi(n)=\varphi(2^a)\varphi(p_1^{e_1})\varphi(p_2^{e_2})\cdots \varphi(p_k^{e_k}).$$
We find all $n$ such that $\varphi(n)=6$. If $k \ge 2$, then since $\varphi(p_i^{e_i})$ is even, $\varphi(n)$ is divisible by $4$, so cannot be equal to $6$.
If $k=0$, then $\varphi(n)=\varphi(2^a)$. But $\varphi(2^a)=1$ if $a=0$ and $\varphi(2^a)=2^{a-1}$ if $a \ge 1$. So if $k=0$ we cannot have $\varphi(n)=6$.
We conclude that $k=1$. Thus $n$ must have the shape $2^ap^e$, where $a \ge 0$ and $p$ is an odd prime.
But $\varphi(p^e)=p^{e-1}(p-1)$. It follows that $p \le 7$. If $p=7$, then $p-1=6$, so we must have $e=1$ and $\varphi(2^a)=1$. This gives the solutions $n=7$ and $n=14$.
We cannot have $p=5$, for $4$ divides $\varphi(5^e)$.
Let $p=3$. If $e \ge 3$, then $\varphi(3^e)\ge (3^2)(2)$. So we are left with the possibilities $e=1$ and $e=2$.
If $e=1$, then $\varphi(n)=\varphi(2^a)(2)$. This cannot be $6$.
Finally, we deal with the case $e=2$. Note that $\varphi(3^2)=6$. So to have $\varphi(2^a3^2)=6$, we need $\varphi(2^a)=1$, which gives $n=9$ and $n=18$.
As you've noted, if $m\ge4$ the sum is $3$ more than a multiple of $5$. You're right, that can't be square: the only allowed residues are $0,\,1$.
Best Answer
This is a finite problem. Since $x,y>0$ we must have $y>x$. But $(x+1)^2-x^2=2x+1$ so $3y^2-3x^2≥6x+3$.
It follows that $$6x+3≤2x+92\iff 4x≤89\iff x≤22$$
A simple search then confirms that the OP has found the only two solutions: $(x,y)=(2,6)$ and $(x,y)=(8,10)$.