Solution: n=$2*3^a$ k=$3^a$

Or just take $k>3$ any prime, and $n$ such that $n\equiv 1\pmod{p-1}$ and $n\equiv 3\pmod{p}$ (such $n$ exists due to CRT). By Fermat, we are done.

I think you left out $n>k$, otherwise all $k$ works by letting $n=1$.