58th Ukrainian National Mathematical Olympiad

Without loss of generality let $z\ge \max \{x,y\}$. Consider the following cases.

Case 1. $z>1$. Then due to the first equation $x+y^2\le z+z^2<2z^3$ – contradiction.

Case 2. $z=1$. Due to the first equation, $2=x+y^2$. Since $\max \{x,y\}\le 1$, it is possible only if $x=y=1$. It is easy to check that $(1;1;1)$ is a solution.

Case 3. $z<1$. Then for the smallest variable, say $x$, $x<1$ holds, then by the second equation: $x+y^2\le z+z^2<2z^3$ – contradiction.

Similar contradiction holds if $y$ is the smallest variable.