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

Write
$10!
= 10.9.8.7.6.5.4.3.2
=2^83^45^27
$.
You want
$(x+y)(x-y)=10!=ab$
with
$x+y=a$ and $x-y=b$.
Then
$x = (a+b)/2$ and
$y = (a-b)/2$.
So $a$ and $b$
have to have the same parity.
Then choose $2^a3^b5^c7^d$
with all possible choices
satisfying the above.
Since they have to
have the same parity,
$a$ must have
$2^n$ where
$1 \le n \le 6$.
This is a start.
