Solution:
The given equation can be written into the form
$$2(x+y)^{2}+(x-y)^{2}=664$$
Therefore, both numbers $x+y$ and $x-y$ are even.
Let $x+y=2 m$ and $x-y=2 t$, $t \in \mathbb{Z}$.
Now from (1) we have that $t$ and $t^{2}$ are even and $m$ is odd.
So, if $t=2 k$, $k \in \mathbb{Z}$ and $m=2 n+1$, $n \in \mathbb{N}$, then from (1) we get
$$k^{2}=41-2 n(n+1)$$
Thus $41-2 n(n+1) \geq 0$ or $2 n^{2}+2 n-41 \leq 0$. The last inequality is satisfied for the positive integers $n=1,2,3,4$ and for $n=0$.
However, only for $n=4$, equation (2) gives a perfect square $k^{2}=1 \Leftrightarrow k= \pm 1$. Therefore the solutions are $(x, y)=(11,7)$ or $(x, y)=(7,11)$.