After rearranging we get: $(k-n)(k+n) = 3^n$
Let $k-n = 3^a, k+n = 3^{n-a}$
we get: $2n = 3^a(3^{n-2a} - 1)$ or, $(2n/(3^a)) + 1 = 3^{n-2a}$
Now, it is clear from above that $3^a$ divides $n$ . so, $n \geq 3^a$
If $n = 3^a, n - 2a = 3^a - 2a \geq 1$ so $RHS \geq 3$ But $LHS = 3$
If $n > 3^a$ then $RHS$ increases exponentially compared to $LHS$ so $n$ cannot be $> 3^a$ .
Thus $n = 3^a$ .
Substituting value of $n$ above we get:
$3 = 3^{3^a - 2a}$
or $3^a - 2a = 1$ this results in only $a = 0$ or $a = 1$
Thus $n = 1$ or $3$ .
~Kris17