Prove 5^n + 3 <> 2^m for all n > 3.

Express 5 as (4+1) and expand using the binomial theorem.
We define p(n,r) = n!/((n-r)!r!) as the binomial coefficients
(1) (4+1)^n = 2^m - 3
The first n-1 terms of the b-expansion is a multiple of 4^(n-r) with the n-1 st term is exactly 4* an odd number. Then upon expansion of (1) we have
4^n + p(n,1)*4^(n-1)+p(n,2)*4^(n-2)+...p(n,r)*4^(n-r)+ 1 =   2^m - 3
adding 3 we get
4^n + p(n,1)*4^(n-1)+p(n,2)*4^(n-2)+...p(n,r)*4^(n-r)+ 4 = 2^m
dividing out 4 we have
4^n-1 + p(n,1)*4^(n-2)+p(n,2)*4^(n-3)+...p(n,r) + 1 = 2^(m-2)
This implies that p(n,r) is odd and therefore n must be odd also. Now p(n,r)+1 must be of the form 2^q - 1 to get the max value of m for large n. This is a start I think.

k = 2^(m-2) - 1
Therefore n must be odd.