I have found what appears to me to be an elementary argument that Catalan's Conjecture is true in the case where $x,y$ are prime.
Please let me know if I have made a mistake.
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 $x=3,a=2,y=2,b=3$
In a previous question, I attempted to prove the case where:
$x=2, y$ is an odd prime
To complete the proof for the prime version of the Catalan conjecture, I will attempt to prove the case where:
$x$ is an odd prime, $y=2$
(1) For such a solution:
$$2^b = x^a - 1 = (x-1)(x^{a-1}+x^{a-2}+\dots + x + 1)$$
(2) Both $(x-1)$ and $(x^{a-1} + \dots +1)$ are powers of $2$ so that there exists $h$ such that:
$x-1 = 2^h$
$(x^{a-1} + \dots + 1) = 2^{b-h}$
(3) There is no solution if $h > 1$ since:
If $h > 1$, then $x \ge 5$ and $x+1$ is not a power of $2$.
$a-1$ must be odd since $(x^{a-1}+x^{a-2}+\dots + x + 1)$ is even.
$(x^a - 1)$ cannot be a power of $2$ since $x+1$ divides $(x^{a-1}+x^{a-2}+\dots + x + 1)$ since $(x^{a-1}+x^{a-2}+\dots + x + 1) \equiv (-1 + 1) + \dots + (-1+1) \equiv 0 \pmod {x+1}$
(4) There is no solution for $h=1$ where $a > 2$ since:
From step #3, $a$ is even so that we have: $2^b=3^a - 1$ = $(3^{\frac{a}{2}} - 1)(3^{\frac{a}{2}} + 1)$
But for $a > 2$, $\frac{a}{2} \ge 2$ and either $3^{\frac{a}{2}} - 1$ or $3^{\frac{a}{2}} + 1$ is not a power of $2$.
Edit: Attempting to greatly simplify the argument based on feedback received.