CroatianCompetitions

The given equation is equivalent to
$$\begin{aligned} x^2y - x^2 + xy^2 - y^2 &= 1 \\ xy(x + y) - (x^2 + y^2) &= 1 \\ xy(x + y) - ((x + y)^2 - 2xy) &= 1 \end{aligned}$$
If we put $u = x + y$ and $v = xy$ we get the equation:
$$uv - (u^2 - 2v) = 1$$
from which follows $uv + 2v = u^2 + 1$, that is
$$v = \frac{u^2 + 1}{u + 2} = \frac{u^2 - 4 + 5}{u + 2} = u - 2 + \frac{5}{u + 2}.$$
Now we can see that $u + 2$ has to be a divisor of $5$, so we have four cases:
$$\begin{array}{lcll} u + 2 = 5 & u + 2 = 1 & u + 2 = -1 & u + 2 = -5 \\ \frac{5}{u+2} = 1 & \frac{5}{u+2} = 5 & \frac{5}{u+2} = -5 & \frac{5}{u+2} = -1 \\ u = 3 & u = -1 & u = -3 & u = -7 \\ v = 2 & v = 2 & v = -10 & v = -10 \end{array}$$
From Viète's formulas we know that $x, y$ are the solutions of quadratic equations:
$$\begin{aligned} z^2 - 3z + 2 &= 0 \\ z^2 + z + 2 &= 0 \\ z^2 + 3z - 10 &= 0 \\ z^2 + 7z - 10 &= 0 \end{aligned}$$
The solutions of the first equation are $1$ and $2$. The solutions of the second equation are not real because its discriminant is negative. The third equation has solutions $-5$ and $2$, while the solutions of the fourth equation are irrational because its discriminant $89$ is not a perfect square.
Finally, the required solutions are $(x, y) \in \{(1, 2), (2, 1), (2, -5), (-5, 2)\}$.