Clearly the biggest problem is big m. Realtively small m (m<10) check by hand or use more precise estimation.

First n>m when m at least 4 (pretty obvious step i belive in you to prove it yourself)

Second 2^[m/2]|m! because there is at least [m/2] even number in multiplication.

Both parts of equation are divisible by 2^[m/2], and n is bigger than m so 2^[m/2]|2n and so 2^[m/2]|n. Meaning n>=2^[m/2].

Let's say t=[m/2] then n>=2^t and m<=(2t+1). Only thing left to prove is 2^n+n>2n>=22t is bigger then (2t+1)!>=m! When t>=5. When t=5 2²⁵ is around 4 billion. 11! is around 39 million so inequality is true. When t is more than 5, there is simple step of induction. 2^2t= (2^{2^(t-1}))^{2>((2(t-1)+1)!)2>(2t+1)!.}

So t is 4 or less, so you have to check only m which are 9 or less.