Let us introduce positive integer $k$ such that $2^k=\left[\frac{2^n}{n}\right ]$. If $n=4$ there is nothing to prove, so let us assume $n>4$. One can easily check inequality $2^n>n^2$, so $2^{2k}\geq \frac{2^{2n}}{n^2}>2^n$ or $k>2n$. Therefore $2^n+2^{n-k}-2^k-1<2^n$ and $2^{n-k}-1<\frac{2^n}{2^k+1}<n\leq \frac{2^n}{2^k}=2^{n-k},$ so $n$ is situated between two consequtive integers.It is possible iff $n=2^{n-k}$, q.e.d.