Find all $x \in \mathbb Z$ and $y \in \mathbb N$ such that $x^2-y!=2001$ and prove your answer is correct

Question

I have the following question:

Find all $x \in \mathbb Z$ and $y \in \mathbb N$ such that $x^2-y!=2001$. Prove that your answer is correct.

And I have absolutely no idea where to even start. I don't even know what my end going is going to be here. What do I have to do first?

Answer 1

For large enough $y$, $y!$ will be divisible by $9$ and then $x^2=y!+2001\equiv3\pmod 9$ which is impossible. How large, is large enough?

Answer 2

By trial of the first $5$ values for $y$

$$(y,x)=(4,\pm45)$$

For $x>5$, see previous answer.