If we consider $k \ge n$ then we need $$4^n \mid 3^k - 1 \qquad \text{and} \qquad 3^n \mid 4^k - 1.$$By Euler-Fermat we can thus take any $k \ge n$ which is divisible by both $\phi ( 3^n)$ and $\phi( 4^n)$.