If $z=0$ we must have $x=y=0$, now assume $z\geq1$. Observe that the first equation is equivalent to $(zx-1)(zy-1)=z^3+1$, and since $z\leq\min(x,y)$, we get\[(z^2-1)^2\leq(zx-1)(zy-1)=z^3+1\iff z^4-z^3-2z^2\leq 0\iff (z-2)(z+1)\leq0.\]This means $z\in\{1,2\}$. If $z=1$, we have $(x-1)(y-1)=2$, from which we get the pairs $(x,y)=(2,3), (3,2)$. If $z=2$, we have $(2x-1)(2y-1)=9$, from which we get $(x,y)=(2,2)$. Hence, the solutions for $(x,y,z)$ are $\boxed{(0,0,0),(2,3,1),(3,2,1),(2,2,2)}$. It is easy to check all four satisfy the system.

If $z=0$ then $x=y=0$ Assume that $x\ge y \ge z$ $$x^2 +2x \ge x+y+z^2=xyz$$ $$\implies x+2\ge yz$$ $$y+z^2=x(yz-1)\ge (yz-2)(yz-1)=(yz)^2-3yz+2$$ $$0 \ge z^2(x^2-1)-z(3x)+2-x$$ it is obvious that $y>1$ assume the function $$f(z)= z^2(x^2-1)-z(3x)+2-x$$$x\ge2$ it is easy to get that f(z) is increase if $z\ge 3$ $$f(z)\ge 9(x^2-1)-9x+2-x=9x^2-10x-7\ge 0$$if$z=2$ $$x+y+4=2xy\implies (2x-1)(2y-1)=9$$ $$\implies x=y=2 or y=1$$ wrong because $y\ge z$ if$z=1$ $$x+y+1=xy \implies (x-1)(y-1)=2$$ $$\implies x=3,y=2$$and we are proved