We immediately see that $n = 1$ does not lead to a solution, while $n = 2$ yields the solution $(k, n) = (45, 2)$.

We show that there is no solution with $n \ge 3$. In that case $3^n$ is divisible by $9$ and thus $k^2$ is divisible by $9$ which implies that $k = 3l$ for some positive integer $l$. After division by $9$, the equation reads $l^2 - 224 = 3^{n-2}$. Modulo $3$, this yields $l^2 - 2 \equiv 0 \pmod{3}$, a contradiction because $2$ is not a quadratic residue modulo $3$.