Does every odd integer $m$ satisfy $3^x(m)-2^y=1$ for some integer values of $x$ and $y$? 
Does the equation $$3^x(m)-2^y=1$$ have positive integer solutions $x, y$ for for every positive odd number $m$?

For example, for $m = 1$, we have $x = 1, y = 1$: $3^1(1)-2^1=1$. For $m=3$, the (only) solution is $x=1,y=3$. But what about the general case?
This question looks like Mihăilescu's theorem, which proves that the only solution to $3^x-2^y=1$ is $x=2$ and $y=3$, but of course we have the extra multiplicand m in there, and what I want to prove is in fact that there are (or aren't) solutions for all positive odd numbers m.
I've been looking into an unrelated problem and it would be helpful to prove or disprove this but I really don't know where to start. My inclination is to say that there must be solutions $x,y$ for all $m$, because with an infinite number of powers of two and an infinite number of powers of three to work with there will always be a pair somewhere that will have the necessary relation to one another. But I'm lost as to how to translate that into proof, if indeed the statement is even true.
Any help - even partial help - would be greatly appreciated.
Edit: Thanks Travis, thanks Conrad, that solves it for me. I think I can't accept either of you as the "solution" here (I'm new!) but tell me if that's untrue. And thanks!

No, take $m$ to be a power of $3$, your question reduces to:
Distance between powers of 2 and 3
Hope it helps:)
