Austrian Mathematical Olympiad

Since $(x - 1) \cdot x \cdot (x + 1) + (y - 1) \cdot y \cdot (y + 1) = x^3 + y^3 - x - y$, adding $3xy(x + y)$ to both sides of the equation yields the equivalent equation
$$(x+y)^3 - (x+y) = 24 + 3xy(x+y-3) \iff (x+y)^3 - 27 - (x+y-3) = 3xy(x+y-3).$$
Since $(x + y)^3 - 27 = (x + y - 3)((x + y)^2 + 3(x + y) + 9)$, this is equivalent to
$$(x+y-3)((x+y)^2+3(x+y)+9-1-3xy) = 0 \iff (x+y-3)(x^2-xy+y^2+3x+3y+8) = 0.$$
If the expression $x + y - 3$ is equal to $0$, we obtain the set of solutions
$$\{(t, 3-t);\ t \in \mathbb{Z}\}.$$
It remains to find all solutions of the equation $x^2 - xy + y^2 + 3x + 3y + 8 = 0$.
If we consider the equivalent equation $x^2 - (y-3) \cdot x + y^2 + 3y + 8 = 0$ as a quadratic equation in $x$, the discriminant $(y-3)^2 - 4(y^2 + 3y + 8) = -3y^2 - 18y - 23 = 4 - 3(y+3)^2$

must be a perfect square if the solutions are to be integers. This is the case iff $(y+3)^2 = 0 \lor (y+3)^2 = 1$, i.e. iff $y = -2 \lor y = -3 \lor y = -4$.
For $y = -2$ we obtain the equation $x^2 + 5x + 6 = 0$ for $x$, and thus the solutions $(-2, -2)$ and $(-3, -2)$.
For $y = -3$ we obtain the equation $x^2 + 6x + 8 = 0$ for $x$, and thus the solutions $(-2, -3)$ and $(-4, -3)$.
Finally, for $y = -4$ we obtain the equation $x^2+7x+12=0$ for $x$, and thus the solutions $(-3, -4)$ and $(-4, -4)$, completing the set of solutions. $\square$