Searched to 10^8. If this sequence is finite with 34 terms, then there is only one solution to m! + 9 = n^2: m=6, n=27.
Consider the Diophantine equations m^2 - 9 = k*(n^2 - 9), which lead to the Pell-like equations m^2 - k*n^2 = 9*(-k + 1) (1) where the only primes dividing k are in P = {2,3,5,7,11,13,17,19} and such that for every k there is at least one solution (m,n) that belongs to the sequence.
As an example, if k = 10 = 2*5, the 2 pairs of solutions of m^2 - 10n^2 = -81; (27,9) and (333,1053) belong to the sequence but the other solutions seem to not be P-smooth. If k = 30 = 2*3*5; m^2 - 30n^2 = -261 and (147, 27) is a solution belonging to the sequence, ...
If the infinitely many solutions of the Pell-like equations are never P-smooth, then this sequence is finite and there is a precise answer to the extended Brocard's problem: There are exactly 3 solutions to m!+t^2 = n^2 with t=1; 1 solution with t = 3; ... -
Robin Garcia, Oct 01 2012
All terms beyond the first are 3 mod 6, since otherwise n^2 - 9 = 2^k for some k and so n - 3 and n + 3 are both powers of 2. For n > 7 all terms are +-3 mod 30, a consequence of Størmer's theorem. -
Charles R Greathouse IV, Oct 01 2012
