First, let's find two obvious solutions. Suppose that $x=y=z$. Then, this reduces to $x^2=2x$, which has solutions $0$ and $2$, meaning two possibilities of $(0,0,0)$ and $(2,2,2)$. Now, we look for other solutions. The second equation can be rearranged to get $z=y^2-x$, and substituting this into the first equation gives $x^2=y+y^2-x$, or $x^2+x=y^2+y$. With similar logic on the second and third equations, one can show that $z^2+z$ must be equal to these two quantities as well. Let this quantity be $c$. Then $t^2+t=c$ can have at most two solutions, so we conclude that two of $x$, $y$, and $z$ must be equal. Without loss of generality, suppose $x=y$. Then our system reduces to $x^2=x+z$ and $z^2=2x$. Multiplying both sides of the first equation by $4$ gives $(2x)^2=4x+4z$, and substituting the second equation in gives $z^4=2z^2+z$, or $z^4-2z^2-4z=0$. We already know from the earlier solutions that $z=0$ and $z=2$ are two solutions for $z$. The equation $z^2=2x$ tells us that these values of $z$ will give the same equal values for $x$ and $y$, and since we have already found those solutions, we no longer need be concerned with them. Dividing through by $z(z-2)$ gives $z^2+2z+2=0$, which has solutions $z = -1 + i$ and $z = -1 - i$. Taking $z = -1 + i$ gives the solution $(-i, -i, -1+i)$, while $z = -1 -i$ gives $(i, i, -1-i)$. We can check that both of these solutions do indeed work. Of course, we assumed that $x$ and $y$ are equal, so permutations of these are vaid solutions as well. We are finished. In total, there are eight solutions: $(0,0,0)$, $(2,2,2)$, $(-1+i,-i,-i)$, $(-i,-1+i,-i)$, $(-i,-i,-1+i)$, $(-1-i,i,i)$, $(i,-1-i,i)$, and $(i,i,-1-i)$. $\boxed{}$