The trivial answer: yes, there is at least one couple : (11, 12); it creates the same squares as (11, 12).
On a broader note let us address the absolute values and look for additional couples.
Since 10! is over 1,000,000 and 9!+8! below 1,000,000 we need not to check candidates overv 9! . The difference between adjacent squares is always odd and all the factorials, except 1!, are even – therefore one of the couples must include the number 1.
We are looking for n such that n!+1 is a square: it works for n in (4,5,7 )==>(25,121,5041).
Checking for neighbors of 71, i.e. 70 or 72 creates no solution.
Both 5 and 11 lead to the single (in absolute values) i.e.
1!+5! = 11^2
4!+5! = 12^2
BTW, if a 3way representation were allowed:
1!+7! = 71^2
4!+5!+7! = 72^2
