A special case of Catalan's conjecture Solve equation
$$ y^p -  (2^p-1)^x = 1 $$
where $x,y>0 \in \mathbb{Z}$, $p \in \mathbb{P}$. 
Is there a elementary method to do it?
Thanks. =)
 A: Suppose the equation is
$$y^p-z^r=1,$$
with $p,r$ odd primes. The classical approach to Catalan's conjecture was to consider two cases (similar to Fermat's last theorem) which go as follows:
First you rearrange the equation as
$$(y-1)\left(\frac{y^p-1}{y-1}\right)=z^r$$
and then you consider the $\gcd$ of the factors on the left, it can only take the values $1$ or $p$. The first case $\gcd(y-1,\frac{y^p-1}{y-1})=1$ was shown to have no solutions by Cassels

J.W.S. Cassels, On the equation $a^x-b^y=1$, II, Proc. Cambridge Philos. Soc. 56 (1960), 97-103

with another proof given by S. Hyyro later. Cassels' proof uses elementary techniques. The punchline is that the second (hard) case is when $r| y$ and $p|z$.
Coming back to your equation we see that $2^p-1\equiv 1 \pmod{p}$, so we are in the first case, and you do not need the full strength of Mihăilescu's proof.
A: I think this gives a nice elementary solution, too.
See this: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=505773
