working in modulo 3 we have
$ 615 \equiv 0 \pmod 3$
as LHS is 0/1 mod 3 so we must have RHS 1 mod 3(as it cannot be 0 mod 3)
so n has to be even (as 2^n is 1 mod 3 for n even) so n = 2m
so
$x^2+615 = 2^{2m}$ or $2^{2m}- x^2 = 615$ or $(2^m + x)(2^m - x) = 615$
so we have the following combinations (615,1) , (205,3), (123,5), (41,15), out of which we get solutions $(2^m= 64,x= 59)$, others are not power of 2 so $(x=59, n=12)$ is the solution
