The Problems of Ukrainian Authors

Note that if $(2^k)!:2^l$, then the pair $(l,(2^k)!)$ satisfies the equation and vice versa: if a pair $(n,m)$ is the solution of equation, then $(2^k)!:2^n$. That is, the number of solutions equals to the number of divisors of the kind $2^l$ of $(2^k)!$. By the Legendre theorem the latter equals: $[2^k/2]+[2^k/2^2]+[2^k/2^3]+\dots =2^{k-1}+2^{k-2}+\dots +2^1+2^0=2^k-1$.