It is not too hard to show that there are no solutions. The assumption that $2^k-3^n$ divides $3^n-2^n$ implies that the former quantity divides $2^{k-n}-1$ and, further, that $2^k < 3^{n+1}$, so that $k < (n+1) \log (3)/\log(2)$. We thus have $$ \left| 2^k - 3^n \right| < 2^{(n+1) \log (3)/\log(2)-n}. $$ On the other hand, standard lower bounds for linear forms in $2$ logarithms show that $$ \left| 2^k - 3^n \right| \geq \min \left\{ 2^k, 3^n \right\}^{0.9}, $$ say, with precisely $23$ exceptions (this is a result from de Weger's thesis; the largest exception is with $(k,n)=(84,53)$). Putting these inequalities together gives you what you want.