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.

Guide:
Case $1$: If $z=1$. Check what happens here.
Case $2$: If $z \ne 1$, then $x+y \ne 0$,
$$x^2-xy+y^2=1+z=1+(1-(x+y))$$
$$x^2+y^2-xy+x+y = 2$$
$$(x^2-xy+x)+(y^2+y)=2$$
$$(x^2-x(y-1))+(y^2+y)=2$$
$$\left(x - \frac{y-1}2\right)^2-\left(\frac{y-1}2 \right)^2 + (y^2+y)=2$$
$$(2x-y+1)^2 -(y^2-2y+1) + 4y^2+4y=8$$
$$(2x-y+1)^2 + 3y^2+6y-1=8$$
$$(2x-y+1)^2+3(y^2+2y+1)=12$$
$$(2x-y+1)^2+3(y+1)^2=12$$
Hence we have $|y+1| \in \{1,2\}$.

*

*If $|y+1|=1$, then $|2x-y+1|=3.$

*If $|y+1|=2$, then $|2x-y+1|=0.$
I will leave the rest as an exercise.
