Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$. Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.

I think we'll have to use number theory to do it. Simply solving the equations won't do.
If we divide the second equation by the first, we get:
$$x^2 - xy + y^2 = 1 + z.$$
Also, since they are integers $z^2 \ge z \implies -z^2 \le -z$. This implies
$$x + y = 1 - z \ge 1 - z^2 = x^3 + y^3.$$
This shows that atleast one of $x$ and $y$ is negative with the additive inverse of the negative being larger than that of the positive.
I have tried but am not able to proceed further. Can you help me with this?
Thanks.
A general identity that is nice to know is
$$x^3+y^3=(x+y)(x^2-xy+y^2).$$
Given that $x+y=1-z$ and $x^3+y^3=1-z^2$ we have $z=1-x-y$ and hence
$$x^3+y^3=1-z^2=(1-z)(1+z)=(x+y)(2-x-y).$$
This means that either $x+y=0$, so $y=-x$ and $z=1$, or
$$x^2-xy+y^2=2-x-y.$$
This is a quadratic in $x$, and quadratics are easy. Applying the quadratic formula yields
$$x=\frac{y-1\pm\sqrt{(1-y)^2-4(y^2+y-2)}}{2}=\frac{1}{2}\left(y-1\pm\sqrt{-3(y+3)(y-1)}\right),$$
and for these $x$ to be integers we need $-3(y+3)(y-1)$ to be a square, so certainly this must be nonnegative. It follows that 
$$y\in\{-3,-2,-1,0,1\},$$
and these few cases can be checked by hand. For $(x,y,z)$ we find the solutions
$$(-2,-3,6),\ (-3,-2,6),\ (0,-2,3),\ (-2,0,3),\ (1,0,3),\ (0,1,3),$$
and $(x,-x,1)$ for all $x\in\Bbb{Z}$ from before.
$$y=-x, z=1$$

$${y = \frac{-1 + x - \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x + \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$

$${y = \frac{-1 + x + \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x - \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$

All integer solutions:

$x=-3,   y=-2,   z=6$

$x=-2,   y=-3,   z=6$

$x=-2,   y=0,   z=3$

$x=0,   y=-2,   z=3$

$x=0,   y=1,   z=0$