This is a complicated way of asking to solve $ n!=2^{m}m!$ for $ m\ge2$. Looking at the power of $ 3$ in both sides, we see that $ n<m+3$ (otherwise $ m!$ has a factor 3 more than the $ n!$), so either $ n=m+1$ (then $ m+1=2^{m}$ implies $ m=1$) or $ n=m+2$ (then $ (m+1)(m+2)=2^{m}$ implies $ m=0$). So no solutions.

Firstly $ n>m$, sof $ n=m$ or $ n=m+1$ and no more. That's $ m+1=2^{m}$ which is simply impossible. The same method, we could solve the equation: $ n!\cdot k!=p^{m}m!$ where $ p$ is prime.