Already discussed before (especially not new)...

Easy to check the equation not have solution (x,y,z) where all $ x,y,z$ different from 0 Case: x=0 $ y^{2}= z^{4}$ Case 2 y=0 $ x^{4}= y^{4}$

I think it appear in DIOPHANTE EQUATION,A book of Dr.ANDREESCU

Why is this very famous problem in an Olympiad Cotest?

We will prove that the only solutions are ones where $xyz = 0$ (it's easy to solve the problem in this case). Suppose, for contradiction, that there existed a solution where $x, y, z > 0$. Take the one where $z$ is minimized. Then we clearly have that $x, y, z$ are relatively prime. We consider two cases. Case 1. $z, x$ are both odd. Then we know that $z^2 - x^2 = 2a^2, z^2 + x^2 = 2b^2$ for some $a, b \in \mathbb{N}.$ This means that $z^2 = a^2 + b^2$ and $x^2 = b^2 - a^2$, so that $a^4 + (zx)^2 = b^4.$ However, $b < z$ and so this contradicts the minimality of $z.$ Case 2. $z, x$ are not both odd. Then we must have that $z$ is odd and $x$ is even. We have two subcases here. In the first case, there are integers $a, b \in \mathbb{N}$ so that $a^2 + b^2 = z^2$ and $a^2 - b^2 = x^2$. Then we have $a^4 - b^4 = (xz)^2$ and $a < z$, contradicting the minimality of $z$. Else there are integers $a, b \in \mathbb{N}$ so that $a^2 + b^2 = z^2$ and $2ab = x^2.$ Note that $\gcd(a, b) = 1$. The first equation yields that there are integers $t, u \in \mathbb{N}$ so that $\{a, b\} = \{t^2 - u^2, 2tu\}$. This means that $x^2 = 2 \cdot (t^2 - u^2) \cdot 2tu$, and so $tu (t^2 - u^2)$ is a square. We have that $t, u, t^2 - u^2$ are pairwise relatively prime and so they're all squares. Hence, $(\sqrt{t})^4 - (\sqrt{u})^4$ is a square. As $\sqrt{t} < z$, this contradicts the minimality of $z$. We've derived a contradiction in all cases, and so our assumption that there is a solution where $xyz \neq 0$ was incorrect. So all solutions are those where $xyz = 0$ (easy to find). $\square$