[seqfan] Double factorial analogue of Brown Numbers

[Jonathan Post]

Brown Numbers
http://mathworld.wolfram.com/BrownNumbers.html

Brown numbers are pairs (m,n) of integers satisfying the condition of
Brocard's problem, i.e., such that
n! + 1 = m^2

where n! is the factorial and m^2 is a square number. Only three such
pairs of numbers are known: (5, 4), (11, 5), (71, 7), and Erdős
conjectured that these are the only three such pairs.

If we use , instead of n!, A001147 Double factorial of odd numbers:
(2*n-1)!! = 1*3*5*...*(2*n-1).

We see that (2,2) and (3,4) are solutions.

(2*2-1)!! = 1*3 = 3 and 3+1 = 4 = 2^2
and
(2*3-1)!! = 1*3*5 = 15 and 15+1 = 16 = 4^2.

Are there more solutions?

What if we use n!!! or n!!!! or others in OEIS?

REFERENCES:
Guy, R. K. Unsolved Problems in Number Theory, 2nd ed. New York:
Springer-Verlag, p. 193, 1994.
Pickover, C. A. Keys to Infinity. New York: Wiley, p. 170, 1995.