Well, if you would have $n|x^2-2$, then putting $y = \frac{x^2-2}{2}$ gives us a legitimate solution. So we only need to be worried about solutions to that divisibility.
You can easily see, that if $x$ is a solution, then $n-x$ and $n+x$ are also solutions. That yields your observation with summing up to $n$.
Now we will focus on the set $S$ of such $n$, that some solution exists.
If $k$ and $l$ are coprime numbers in $S$, then using Chinese Remainder Theorem we can find solution for $kl$. Moreover, the number of solutions for $kl$ will be equal to the product of these numbers for $k$ and $l$. We also see quite obviously that if $n \in S$, then all divisors of $n$ are in $S$.
From these two facts, we only need to be worried about $n$ equal to prime powers. It is not very easy to prove that $n=p^k$ works iff $n=2$ or $p \equiv 1,7 \pmod 8$, but you should be able to find proof of this fact in any materials about quadratic residues (here, for example, $2$ is a quadratic residue mod $n$, since $x^2 \equiv 2 \pmod n$).
So, solutions exist for $n$ iff all of its prime divisors are $2$ or primes $\equiv 1$ or $7 \pmod 8$, where $2$ can show up only once (so $n$ can't be divisible by $4$, but it can be even). Number of these solutions (or, to be exact, number of different families of sulutions modulo $n$) is equal to $2^k$, where $k$ is number of distinct odd prime divisors of $n$.
Best Answer
Noting that the coefficient of the middle term is the sum of the other coefficients, so there is a factor $x+y$, we find the factorisation $y=(x+y)(3x+4y)$
Set $x+y=X$ and we have $y=X(3X+y)$ and $y=\cfrac {3X^2}{1-X}$
($X=1$ is easily eliminated, so we are not dividing by zero)
Now $X$ and $1-X$ can have no [non-trivial] factor in common, so $1-X$ must be a factor of $3$ hence $(1-X)\in \{-3, -1, 1, 3\}$
Clearly each of these possibilities does provide a solution, since the steps are reversible.