Partial Simplified Proof for the prime version of the Catalan Conjecture I have found an elementary approach which seems to work in proving that Catalan's conjecture is true for specific primes.  I am attempting to generalize this approach to see how far it will go.
I am presenting it here with the hopes of finding a mistake or confirming that this approach is worth exploring further.
From Wikipedia, the Catalan Conjecture can be defined as:


*

*Let $a>1,b>1,x>0,y>0$


*Then if:
$x^a - y^b = 1$

*

*The only solution is $x=3, a=2, y=2, b=3$

I will attempt now to prove the case where:

$x=2, y$ is an odd prime

(1)  Making this assumption, we have:

$$2^a = 2\left[y(y^{b-2} + y^{b-3} + \dots + 1)(\frac{y-1}{2}) + \frac{y+1}{2}\right]$$

(2)  $\frac{y+1}{2}$ is even

If $\frac{y+1}{2}$ is odd, then $\frac{y-1}{2}$ is even and $\left[y(y^{b-2}+\dots+1)(\frac{y-1}{2}) + \frac{y+1}{2}\right]$ is odd.

(3) $b$ is odd

If $\frac{y+1}{2}$ is even, then $\frac{y-1}{2}$ is odd and it follows that $(y^{b-2} + \dots + 1)$ must be even.  Therefore, $b-2$ must be odd.

(4)  Let $2^u$ be the highest power of $2$ that divides $\frac{y+1}{2}$
(5)  $y \equiv -1 \pmod {2^{u+1}}$ since:

There exists $m$ such that $\frac{y+1}{2^{u+1}}=2m+1$ which means $y = 2^{u+1}(2m + 1) - 1$

(6) $2^{u+1} | (y^{b-2} + \dots 1)$ since:

$b-2$ is odd and $(y^{b-2} + y^{b-3} + \dots + y + 1) \equiv (-1 + 1) +\dots + (-1 + 1) \equiv 0 \pmod {2^{u+1}}$

(7)  But then we have a contradiction since:

$y(\frac{y^{b-2}+\dots+1}{2^u})(\frac{y-1}{2})$ is even but $\frac{y+1}{2^{u+1}}$ is odd so that:
$$2^a \ne 2^{u+1}\left[y(\frac{^{b-2}+\dots+1}{2^u})(\frac{y-1}{2}) + \frac{y+1}{2^{u+1}}\right]$$


Edit: Attempting to greatly simplify the argument based on feedback received.

I couldn't find any errors so your approach seems to be correct.
Note, though, that your solution is more general in that it also works for any $x$ being a positive power of $2$, plus for all odd $y \gt 1$, not just odd primes. For the first case, this is because your proof just requires the LHS to be a power of $2$. For the second case, this is due to you only using that $y$ is odd, and in particular not using any particular property of primes, in your solution.
As for "exploring further" with this approach, I'm not quite sure what you mean. However, note you can use a similar approach to show that the only values which work for $y$ being a positive power of $2$ is the only known solution. To show this, you can follow a similar approach to what you did. First, let $y = 2^c$ for an integer $c \ge 1$. Next, the equation can now be rewritten as
$$2^{bc} = x^a - 1 \tag{1}\label{eq1}$$
This shows that $x$ must be odd. First, consider that $a$ is even, i.e., $a = 2d$. Then $x^a - 1 = \left(x^d + 1\right)\left(x^d - 1\right)$, so both $x^d + 1$ and $x^d - 1$ must be powers of $2$.  First, $x^d - 1 = 1 \; \Rightarrow \; x^d = 2$, so it's not possible. Next, $x^d - 1 = 2 \; \Rightarrow \; x^d = 3$, so $x = 3$ and $d = 1$, giving the $1$ known solution. No larger values of $x^d - 1$ as a power of $2$ have that $x^d + 1$ is also a power of $2$.
Next, consider that $a$ is odd. Factoring the LHS gives $x^a - 1 = \left(x - 1\right)\left(x^{a-1} + x^{a-2} + \cdots + x + 1\right)$. Since $x$ is odd and there are $a$ terms in the second factor, this factor must be odd and $\gt 1$. However, the LHS of \eqref{eq1} is a power of $2$, so it has no odd factors. Thus, $a$ can't be an odd integer.
Putting this solution together with what you've shown proves that, apart from the one known solution, there are no other ones where $x$ or $y$ is a power of $2$.
You could try to do something similar with odd prime numbers, but I don't believe that just these sorts of basic remainder and number of factors type arguments are sufficient to show there is no solution if either $x$ or $y$ is of a given simple form (e.g., just a prime or a power of a prime). If you haven't already done so, you may wish to read & analyze Preda Mihăilescu's proof, such as discussed & outlined in the AMS article at Catalan's Conjecture: Another Old Diophantine Problem Solved.
