I hope this is correct: Suppose that at least two of $x,y,z$ are negative. Then, the LHS of some equation is negative, but the RHS is a square, and thus nonnegative; contradiction. Next, suppose that all three are nonnegative. Then, summing the equations, we get $$x^2+y^2+z^2=x^3+x+y^3+y+z^3+z$$but we have, by AM-GM, $$x^2+y^2+z^2=x^3+x+y^3+y+z^3+z\ge 2x^2+2y^2+2z^2$$so we must have $x=y=z=0$. Finally, suppose WLOG (since the system is cyclic) that $z=\text{min}(x,y,z)$, so we have $x,y\ge0\ge z$. Now, let $t=-z\ge 0$. We have $y^3-x^2=t\ge 0$, so $y^3\ge x^2$, and also $x-y^2=t^3\ge0$, so $x\ge y^2$. But, finally, we have $y^3\ge x^2\ge y^4$, so $1\ge y$, from where it follows that $1\ge x$ and $$x^3+y=(y^3-x^2)^2=(y^2)^3-2y^3x^2+x^4\le x^3-2y^3x^2+x^2\le x^3-2y^3x^2+y^3\le x^3-2y^3x^2+y$$Thus, one of $x,y$ must be zero, and we easily get that $x=y=z=0$. So $(x,y,z)=(0,0,0)$ is the only solution.