Find all integers $(x, y)$ such that $x^{2}=2+6 y^{2}+y^{4}$.
As $\left(y^{2}+1\right)^{2}=1+2 y^{2}+y^{4}<2+6 y^{2}+y^{4}<9+6 y^{2}+y^{4}=\left(y^{2}+3\right)^{2}$, and $2+6 y^{2}+y^{4}$ is a perfect square, it must be that $2+6 y^{2}+y^{4}=\left(y^{2}+2\right)^{2}$. We deduce that $2+6 y^{2}+y^{4}=y^{4}+4 y^{2}+4$ which simplifies to $y^{2}=1$ and thus $y= \pm 1$ and $x= \pm\left(y^{2}+2\right)= \pm 3$.

Conversely, we verify that the pairs $( \pm 3, \pm 1)$ work.