Hint

Do with : $8(y+z)^{3}=y^{6}+z^{6}+31(y^{2}+z^{2})$ $min(y,z) \leq 4$

$y$ and $z$ play the same roles , so $y=z$

WLOG $y \ge z$. From $x^6 = y^6 + z^6 + 31(y^2 + z^2)$, it follows that $x^6 > y^6$, hence $x > y$. From $x^2 = 2(y+z)$, it follows $x^2 \le 2(2y) = 4y$. Since $x > y$, then $y^2 < x^2 \le 4y$, hence $y < 4$. Because $z \le y < 4$, it follows that $z < 4$. Hence $y,z \le 3$. Next, since $x^2 = 2(y+z) \le 2(3+3) = 12$ and $2|x^2$, it follows that $x^2 = 4$, hence $y+z = 2$. Hence $y = z = 1$ and thus $x = 2$, which is a solution. Hence the only solution $\boxed{(x,y,z) = (2,1,1)}$.