Integer solutions of $\sum_{j=1}^m\frac{n-m}{j}+\sum_{j=m+1}^n\frac{n}{j}=k$

Question

This is in reference to an answer I gave to this question. I am curious to know if my intuition is correct. Given $n,m\in\mathbb{N}$, with $n>m$, is it possible for $\exp(2\pi in\sum\limits_{j=1}^n\frac{1}{j})=\exp(2\pi im\sum\limits_{j=1}^m\frac{1}{j})$ to hold, other than the solution where $n=2$ and $m=1$? This can be written equivalently as: are there any other solutions $n,m\in\mathbb{N}$, $k\in\mathbb{Z}$ with $n>m$ other than $n=2$, $m=1$ to the equation: $$\sum_{j=1}^m\frac{n-m}{j}+\sum_{j=m+1}^n\frac{n}{j}=k$$ I suspect that the answer is 'no,' and there may be a proof involving residues modulo $j$, though I would be interested in seeing any proof.