Have the equations $y^2=x^3+n!$ and $y^2=x^3-n!$ infinitely many integral solutions?

The Mordell curve $$y^2=x^3+k$$ is known to have finite many solutions for every integer $k\ne 0$. But is it known whether there are infinite many natural numbers $n$ such that $y^2=x^3-n!$ and/or $y^2=x^3+n!$ has an integral point ? In other words :

Have the equations $$y^2=x^3+n!$$ and/or $$y^2=x^3-n!$$ infinitely many integral solutions ?

I checked the "+"-version for $0\le x\le 10^8$ and $1\le n\le 50$ and found the following squares

? for(s=1,50,z=s!;for(n=0,10^8,if(issquare(n^3+z)==1,print(s,"  ",n))))
1  0
1  2
4  1
4  10
4  8158
5  1
6  4
7  1
9  9
15  54180
21  604800
21  2419200
21  7358400
21  9676800
21  16805376
21  25363584
21  67536000
?


For small values $n$, the possible values can be looked up at tables, but for larger $n$ I am not sure how much is known about possible solutions.

• See here : math.stackexchange.com/questions/2125619/… for a similar equation, but probably only finite many (perhaps even only 3?) solutions – Peter Mar 28 '17 at 13:32
• No harm on the question , but what is the motivation behind $n!$? Suppose we replace $n!$ by a simpler function, such as multiples of $n$, or an arithmetic progression, or a subsequence of primes, maybe? If any of these cases have been solved, then we can borrow tools from them? (I have seen Mordell curves, but none of these results). – астон вілла олоф мэллбэрг Mar 28 '17 at 13:38
• For $n=21$, we have several solutions, so why not ask whether there are solutions for even larger $n$ ? – Peter Mar 28 '17 at 13:55
• $32!+2\ 067\ 296\ 774\ 400^3$ is a square!! – Peter Apr 6 '17 at 16:13
• That is excellent. I really like your question, +1. If you want, I can also run code parallel to yours. You must only tell me the algorithm, and what language you are programming in. I am willing to help you. – астон вілла олоф мэллбэрг Apr 7 '17 at 9:07