System of equations $x^3+y=y^2\ \& \ y^3+z=z^2\ \& \ z^3+x=x^2$

If $xyz=0$ so $(x,y,z)=(0,0,0).$

Let $xyz=r\neq0.$

Thus, by the WE Tutorial School's work we obtain: $$(x+1)(y+1)(z+1)=1.$$ or $$x+y+z+xy+xz+yz=-r.$$ In another hand, by your work we obtain: $$(x-1)(y-1)(z-1)=x^2y^2z^2,$$ which gives $$x+y+z-xy-xz-yz=r^2-r+1,$$ which gives $$x+y+z=\frac{(r-1)^2}{2}$$ and $$xy+xz+yz=-\frac{r^2+1}{2}.$$ But after summing of our equations we obtain: $$\sum_{cyc}(x^3-x^2+x)=0$$ and since $$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz$$ and $$x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz),$$ we have: $$\frac{(r-1)^6}{8}+\frac{3(r-1)^2(r^2+1)}{4}+3r-\frac{(r-1)^4}{4}-r^2-1+\frac{(r-1)^2}{2}=0$$ or $$r^6-6r^5+19r^4-24r^3+11r^2+6r+1=0$$ or $$(r^3-3r^2+3r+1)^2+4r^2(r^2-2r+2)=0,$$ which is impossible for real value of $r$.

Id est, $(0,0,0)$ is an unique solution of the system.

I will prove by contradiction that there exists no solutions for $x > 1$ and $y,z < 1$. Suppose $(x,y,z)$ is such a solution.

Iteration #1: First, note that: $$ z^3 = x^2 - x = x(x - 1) > 0 \implies z > 0 $$ Thus, $0 < z < 1$. We see that: $$ y^3 = z^2 - z = z(z - 1) $$ For $0 < z < 1$, $-\frac{1}{4} < z(z - 1) < 0$, so $-2^{-\frac{2}{3}} < y < 0$. For this range of values of $y$, since $y(y - 1)$ is decreasing for $y < 0$, we have that: $$ x^3 = y^2 - y < 2^{-\frac{4}{3}} + 2^{-\frac{2}{3}} = 1.0268108 \implies x < 1.00885823 $$ Iteration #2: Applying this inequality to the third equation again, we have: $$ z^3 = x(x - 1) < 1.00885823(1.00885823 - 1) = 0.0089367 \implies z < 0.20752 $$ For $0 < z < 0.20752$, we have that: $$ 0.20752(0.20752 - 1) < y^3 < 0 \implies -0.54787 < y < 0 $$ Once again, we have: $$ x^3 = y(y - 1) < -0.54787 \left(-0.54787 - 1\right) < 1 \implies x < 1 $$ This contradicts that $x > 1$. Therefore, no non-trivial solutions exist.