Solving for $3^x - 1 = 2^y$ Besides $x=2, y=3$, are there any other solutions? 
I know that if there is another solution:


*

*$y$ is odd since $2^y \equiv -1 \pmod 3$

*$x$ is even since $3^x - 1 \equiv 0 \pmod 8$

*$3 | y$ since $-1 \equiv 2^y \pmod 9$


Are there any other solutions?  If not, what is the argument for showing that if $3^x > 9$, then $2^y \neq 3^x-1$
Thanks,
-Larry
 A: You don't need the full strength of Catalan's conjecture here.
Two solutions are found easily. $3^1-1=2^1$ gives $x=y=1$. $3^2-1=2^3$ gives $x=2$ and $y=3$.
To prove that these are the only solutions, assume $x > 2$. As you have noted, $x$ must be even: $x = 2z$. Now we have
$$
3^x-1 = 3^{2z}-1 = (3^z-1)(3^z+1) = 2^y.
$$
It follows that both $3^z-1$ and $3^z+1$ are powers of $2$. But this is impossible, because both numbers are larger than $2$ (since $z > 1$), and they are only two units apart. QED.
PS: I think I've seen this argument somewhere on math.SE.
A: Catalan's conjecture (now a theorem, with the alternative name "Mihăilescu's theorem") asserts that there are no other possible solutions to the diophantine equation in integers $a,x,b,y > 1$:
$$a^x - b^y = 1$$
other than the solution $(a,b,x,y) = (3,2,2,3)$. As for the proof, I believe that Mihailescu's original proof is not simple enough to be placed into a single answer. Here is a link to a simpler and more elementary one, though it is still considerably long: Elementary proof of Mihailescu's Theorem.
A: This link can help you to get an answer to your question. 
