Macedonian Junior Mathematical Olympiad

It is obvious that $z\ge 0$. If $z\ge 2$, then $x^2+y^4+1\equiv 0(\mathrm{mod}4)$, i.e. $x^2+y^4\equiv 3(\mathrm{mod}4)$. This is not possible because the remainders of squares of integers after division by $4$ are $0$ or $1$. According to that, $0\le z<2$.

If $z=0$, then $x=y=0$.

If $z=1$, then $x^2+y^4=5$, i.e. $(x,y)=\{(2,1),(-2,1),(2,-1),(-2,-1)\}$.

Therefore $(x,y,z)=\{(0,0,0),(2,1,1),(-2,1,1),(2,-1,1),(-2,-1,1)\}$ are the solutions of the given equation.