Find all triples $(x, y, z)$ of positive integers satisfying the system of equations

$$
\left\{\begin{array}{l}
x^{2}=2(y+z) \\
x^{6}=y^{6}+z^{6}+31\left(y^{2}+z^{2}\right)
\end{array}\right.
$$
From the first equation it follows that $x$ is even. The second equation implies $x>y$ and $x>z$. Hence $4 x>2(y+z)=x^{2}$, and therefore $x=2$ and $y+z=2$, so $y=z=1$. It is easy to check that the triple $(2,1,1)$ satisfies the given system of equations.