How many pairs of positive integers $(x,y)$ satisfy the equation $x^2 - 10! = y^2$?

$10!=2^83^45^27$
You want to divide this into two factors, one of which is a multiple of $7$.
The factors must both be even, otherwise $x$ and $y$ will not be integers.
So the factor that includes $7$ must have from one to seven 2's; from zero to four 3's and from zero to two 5's. How many options are there altogether?