Maybe (it's pretty late, but somehow this makes sense at the moment) :
Look at $$ 3^n - 1 = 2^m $$
Factor the difference of two squares on the left, get $$ 3^{\frac {n}{2}} - 1 $$ for one of the factors.
2 must divide it (why?).
It itself is the difference of two squares. Factor it, and keep going.
Eventually, reach 3 - 1. Now this shows that if the original equation holds for any power of 3 greater than 2, one of that form holds for all in that cascade down to 2. Contradict, and conclude.
And if that whole thing is stupid, a brief notice will do, thanks.