Taking the given equation, m2 = (2n + 1)2 + 8·n!, we may try to see if some kind of approximation might help to for large n. Stirling's formula would give, for n >> 1: n! ≈√(2·π·n)·nn·exp(-n). So: m2 ≈ (2n+1)2 + √ [27·π·n2n+1·exp(-2n)]. Then, since m must be integer, m2 should be a square number. We may want to take a few close square numbers, and give then a try with the selected n. Although n= 6 is not much larger than the unit, we may want to try with this number, as other possible integer solutions seem hard to find. For n = 6, last formula yields m2 ≈ 5690. A few close square numbers are: {5476, 5625, 5776, 5929}. It is found that the last number solves the equation; being therefore m = √5929 = 77. Hopefully, for higher n, this kind of search strategy may perform somewhat better, and it might be possible to restrict the set of square numbers to be checked, possibly to just the nearest pair. If a solution is not found for a given n, it seems reasonable to move to next integer. Of course, it would be still quite challenging to deal with arithmetics of very large integers while verifying possible solutions ― if indeed they can be found.

Another way to find a solution, without the need for trying all m’s, could be the following. Let us admit that two solutions could be found: (n, m) and (n + p, m + q). Then both the equations (m+q)2 = (2(n+p) + 1)2 + 8·(n+p)! and m2 = (2n + 1)2 + 8·n! should be valid. Subtraction of the second equation from the first, yields: q·(2m + q) = 4·{p·(2n+1) + p2  + 2·[(n+p)!/n!-1]·n!} q = -m ±2·√{p·(2n+1) + p2  + 2·[(n+p)!/n!-1]·n! + (m/2)2}. It would be now helpful to include possible solutions among no negative integers, so that (n = 0; m = 3) can be considered. Substitution at above equation yields: q = -3 ±2·√{p + p2 + 2·(p!-1) + (3/2)2}. For p = 6 we find: q = -3 ± √5929 = -3 ± 77. We have found a solution: n = 0 + p = 6, m = 3 + q = 3 – 3 + 77 = 77.