I knew I was missing something. I thought of letting m = ab, where a is odd and b is a power of two, and then 3^m-1 = (3^(ab)-1) = (3^a-1)(3^(a(b-1))+3^(a(b-2))+...+1), but the first factor is 2 mod 4, so a must be 1. Then after I posted I looked up a proof and saw that all of the 3^2^k-1 for k>1 are divisible by 3^2+1 = 5*2, and cannot be powers of two. Your proof is much simpler.

I can't prove the fact about Pell numbers, so <student> I will concede the credit there. </student>

I want to continue this, it was funny

Let's say you have 3 wolves and 3 dogs, standing at a side of a river, and they wanna cross the river. They already have a ship, where only 2 animals can fit. BUT if there a MORE wolves than dogs, the wolves eat the dogs! Is there any way to cross the river, and if not, why?

And a bit more harder: is it possible with 4 wolves and 4 dogs?
Good luck!

For the first one:
Carry 2 Wolves across the river, go back.
Carry 2 Dogs across the river, go back.
Now carry the Wolf and the Dog that are left and you are done.

For the second one:
Carry 2 Wolves and go back.
Carry 2 Wolves and go back.
Carry 2 Dogs, carry 2 wolves back.
Carry 2 Dogs and go back.
Carry the two Wolves that are left and you are done .

Sorry, I forgot to say that there's always an animal in the ship...

If you can only fit 2 animals in the ship, there will never be more wolves then dogs (WW,DD,WD,DW) are the only possibilities. Therefore, any combination in any order will work, since you don't have a time limit either.
Answer: Yes there is a way to cross the river, and it is possible to do so with 4 wolves and 4 dogs as well.
I've seen a similar problem with people crossing a bridge in the middle of the night with only 1 flashlight. They had to get across in less than an amount of time. Maybe you've left out a parameter?