Direct evaluation for n=2: 2^(2-2)+2! = 3, which is not power of 2, so n=2 is not a solution.  Then n is odd.

In general, the largest power of 2 that divides n! is given by the sum k={1 to inf} floor[n/2^k]; call that value v.  Since n is odd, let n=2k+1.  Substitute and expand:

v = floor[(2k+1)/2] + floor[(2k+1)/4] + floor[(2k+1)/8] + ....

The +1 component has no effect of the floor[], so we can drop those:

v = floor[k] + floor[k/2] + floor[k/4] + ....

Then we can make an inequality by dropping the floors from the right side expression:

v < k + k/2 + k/4 + ...

Then this is a geometric series, then v < 2k.

But since we are working in the integers, v <= 2k-1 = n-2.

So the power of 2 that divides n! is at most 2^(n-2).  But we need this upper limit to actually be met.

Next I'll compare the two series: floor[n/2] + floor[n/4] + floor[n/8] + ... and n/2 + n/4 + n/8 + ...

Notice that if n is n=2^z, then the two series are identical for the first z terms.  But since n needs to be odd then the closest thing is n=2^z+1.

Let's verify with calculating:

 floor[(2^z+1)/2] + floor[(2^z+1)/4] + floor[(2^z+1)/8] + ...

= 2^(z-1) + 2^(z-2) + ... + 2 + 1 + 0

= 2^z - 1

This is exactly the same as the upper bound n-2=(2^z+1)-2=2^z-1.

So at this point I conclude that n is of the form 2^z+1.  

But we also want n=2^z+1 to be prime, by the problem statement. Assume z=p*q with odd value p and both p and q greater than 1.  Then 2^z+1 = (2^p)^q+1.  But since q is odd then 2^p+1 is a nontrivial factor, and then 2^z+1 is composite. 

So then it is necessary for z=2^y for n=2^z+1 to be prime; then n=2^(2^y)+1. This is the definition of a Fermat Prime. There are only five known Fermat Primes: 3, 5, 17, 257, 65537; and it is conjectured that these are the only such primes.

By calculation only n=3 and n=5 result in a prime m: 2^(3-2)+3!=2^3 and 2^(5-2)+5!=2^7; then the possible pairs (m,n) which satisfy the problem are (3,3) and (7,5).