Find all positive integers $x$ and $y$ such that $x^{2}-y!=2019$.
For $y \geqslant 4, y!$ is divisible by 4, so $x^{2} \equiv 2019 \equiv 3(\bmod 4)$. There is a contradiction because a square is always congruent to 0 or 1 modulo 4. It only remains to handle the cases $y \in\{1,2,3\}$, and we find $x=45, y=3$ as the only solution.