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: $[2^k/2] + [2^k/2^2] + [2^k/2^3] + \ldots = 2^{k-1} + 2^{k-2} + \ldots + 2^1 + 2^0 = 2^k - 1$.