Find all $(x, y) \in \mathbb{Z}^{2}$ such that $x^{3}+2 x y-7=0$.
Let $(x, y)$ be a solution. We factorize: $x\left(x^{2}+2 y\right)=7$. According to the fundamental theorem of arithmetic, we have four cases to distinguish:

- $x=-7$ and $x^{2}+2 y=-1$, so $y=-25$,
- $x=-1$ and $x^{2}+2 y=-7$, so $y=-4$,
- $x=1$ and $x^{2}+2 y=7$, so $y=3$,
- $x=7$ and $x^{2}+2 y=1$, so $y=-24$.

We verify that these four cases indeed give solutions.