How many pairs of positive integers $(x,y)$ satisfy the equation $x^2 - 10! = y^2$? I have a question about number theory:

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

My attempt:
Move the $y^2$ from right to the left and 10! From left to the right such that:
$$x^2-y^2= 10!$$
$$(x-y)(x+y)=10!$$
$$(x-y)(x+y)= 10.9.8.7.6.5.4.3.2$$
Until this step, I think I will get many possibility to get the answer, such that:
$$(x-y)= 10.9$$ and $$(x+y)= 8!$$
Also
$$(x-y)= 10.9.8$$ and $$(x+y)= 7!$$
And another possibilities, like $$(x-y)=10.9.8.7$$ and $$(x+y)=6!$$
And so on until $$(x-y)=10.9.8.7.6.5.4.3.2$$ and $$(x+y)=1$$
And I think solving all the possibility is a bit tedious work. Can somebody help me to find a better solution to solve this kind of problem? 
Thanks

$10!=2^83^45^27$
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?
