This morning, it occurred to me that there's a straight forward argument for Catalan's conjecture for a specific prime and a power of $2$.
I doubt that it will generalize to the full Catalan's conjecture but I am planning to invest time this evening to see what I can come up with.
Is this interesting? Is Catalan's conjecture trivial for a specific prime and a power of $2$?
Has anyone attempted to simplify the argument made by Mihailescu?
Edit: Added an example of the method
Here is a very simple example of the method that I am talking about.
I will use it to argue that there is no solution to $2^a - 5^b = 1$ for any integer $a,b$
Apologies for any typos. Please let me know if you see any mistakes or if any point is not clear.
(1) Assume that such a solution exists.
(2) Then, there exists positive integers $a>0,b>0,x>0,c < 10$ such that:
$2^a = 10x + (c+1)$
$5^b = 10x + c$
(3) Clearly, $c = 5$ so that we have:
$2^a = 10x + 6 = 2(5x + 3)$
$5^b = 10x + 5 = 5(2x + 1)$
(4) We can see that $x$ must be odd otherwise:
$2^a \ne 2(5x+3)$
(5) Further, we can see that $x \equiv 2 \pmod 5$ otherwise:
$5^b \ne 5(2x+1)$
(6) $x \ne 2$ so $x \ge 7$ and there exists $x_2$ such that:
$5^b = 5(2(5x_2+2)+1) = 5(5(2x_2+1))$
(7) We see that $x_2 \equiv 2$ which means that $x = 5x_2 + 2$ but $x_2 \ne 2$ since that would make $x$ even.
(8) We can repeat this argument up until we reach an $x_i = 2$ but then we have the following situation:
$x = 5(5(\dots(5(2)+2)\dots)+2) = (5^{i-1} + 5^{i-2} + \dots + 1)(2)$ which is even.
(9) But this contradicts step #4 where $x$ is odd so we reject our assumption in step #1.