Just from the $k=1$ condition ,we get that $m$ is just a power of 2 or of the form $2^m3^n$, where $m$ and $n$ are positive integers. The first case works obviously as $\varphi(2^k)=2^{k-1}$. Then we investigate the second case . Let $-$ denote an application of phi. Then the second case is a s follows $2^m3^n-2^m3^{n-1}-...2^m3-2^m-2^{m-1}-...2-1$ which also clearly works. Now, just for a more detailed explanation on how the cases appear. We just look at $\frac{m}{\varphi(m)}$ and see when it is an integer.