1. Determine all integer solutions $x, y, z$ such that

$$
x^{2}\left(x^{2}+y\right)=y^{z+1}
$$
Solution. The given equation is equivalent to the equation

$$
\left(2 x^{2}+y\right)^{2}=y^{2}+4 y^{z+1}
$$

It is easily verified that it has no solutions for which $z \leq-1$. For $z=0$, the right side is equal to $y^{2}+4 y=(y+2)^{2}-4$, and can be a square of an integer only for $y=0$ and $y=-4$. The first case gives the solution $(0,0,0)$, and in the second case, there is no solution.

Let us assume that $z \geq 1$. We can rewrite the given equation as

$$
\left(2 x^{2}+y\right)^{2}=y^{2}\left(1+4 y^{z-1}\right)
$$

from which it follows that $1+4 y^{z-1}$ must be a square of an integer. This number must be odd, so $1+4 y^{z-1}=(1+2 v)^{2},(v \geq 0)$, from which we get $y^{z-1}=v(v+1)$. The latter is possible for $v=0$, which gives the solution $x=0, y=0$, and for $z=2$ and $y=v(v+1)$. Substituting into the equation, we get $x^{2}=v^{2}(v+1)$, so it must be $v=t^{2}-1, t \in \mathbb{Z} \backslash\{-1,0,1\}$, (the values $-1,0,1$ are excluded because we again get the trivial solution $x=0$). By direct verification, it is obtained that $x=t^{3}-t, y=t^{4}-t^{2}$ for $z=2$ are solutions to the given equation.

Therefore, all triples $(x, y, z)$ of integers that are solutions to the given equation are $(0,0, z), z \geq 0$ and $\left(t^{3}-t, t^{4}-t^{2}, 2\right), t \in \mathbb{Z} \backslash\{-1,0,1\}$.