Determine all positive integer solutions $x, y$ of the equation $x^{2}-2 \times y!=2021$.
When looking at modulo $8, 2021 \equiv 5$, and $8 \mid 2 \times y$ as soon as $y \geq 4$. Thus, we would have $x^{2} \equiv 5[8]$, which is a contradiction!

Therefore, $y < 4$.

For $0 \leqslant y \leqslant 3$, we get $x^{2}=2023, 2023, 2025, 2030$. Since $44^{2}<2023<2025=45^{2}<2030<46^{2}$, the only possibility is $y=2$ and $x=45$.

Conversely, we verify that the pair $(45,2)$ is a solution.