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

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.