From the first equation we get $x + y = z - 1$. By plugging this into the second equation we get:
$$\begin{aligned} x^2 - y^2 &= 1 - z^2, \\ (x + y)(x - y) &= (1 - z)(1 + z), \\ (z - 1)(x - y) &= -(z - 1)(1 + z), \\ (z - 1)(x - y + z + 1) &= 0. \end{aligned}$$
There are now two cases: $z = 1$ or $x - y + z + 1 = 0$.

If $z = 1$, then $x + y = 0$, i.e. $x = -y$. Plugging this into the third equation we get $2y^3 = -2$, i.e. $y = -1$, $x = 1$.

If $x - y + z + 1 = 0$, then plugging $z = -x + y - 1$ into the first equation yields $2x = -2$, i.e. $x = -1$.
Now we have $z = -1 + y - 1 = y$, which we can combine with the third equation to get $2y^3 = -2$, i.e. $y = z = -1$.

Thus, the possible solutions to the given system of equations are $(1, -1, 1)$ and $(-1, -1, -1)$. A direct computation confirms that both are indeed valid solutions of the system.