How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$? This is INMO 2000 Problem 2.

Solve for integers $x,y,z$: \begin{align}x + y &= 1 - z \\ x^3 + y^3 &= 1 - z^2 . \end{align}

My Progress: A bit of calculation and we get $x^2-xy+y^2=1+z $
Also we have $x^2+2xy+y^2=(1-z)^2 \implies 3xy=(1-z)^2-(1+z)=z(z-3) \implies y=\frac{z(z-3)}{3x}$ and $x=\frac{z(z-3)}{3y} $.
Note that since $z$,$x$,$y$ is an integer, we must have $3\mid z$.
So, let $z=3k$.
So we have $y=\frac{3k(3k-3)}{3x}=\frac{k(3k-3)}{x}$ and $x=\frac{z(z-3)}{3y}=\frac{k(3k-3)}{y}$ .
Then I am not able to proceed.
Hope one can give me some hints and guide me.
Thanks in advance.

We have $(1-z)(x^2-xy+y^2)=1-z^2.$
If $z=1$, so $x+y=0$ and we obtain $(t,-t,1)$, where $t$ is an integer.
Let $z\neq1$.
Thus, $$x^2-xy+y^2=z+1$$ and $$x+y=1-z,$$ which gives $$(1-z)^2-3xy=z+1$$ or
$$3xy=z^2-3z.$$
Thus, $z$ is divisible by $3$ and $$(1-z)^2-\frac{4}{3}(z^2-3z)\geq0$$ or
$$z^2-6z-3\leq0$$ or
$$3-\sqrt{12}\leq z\leq 3+\sqrt{12},$$
which gives $$0\leq z\leq 6$$
Can you end it now?
