Q:

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.

