According to wikipedia’s page on Beal’s Conjecture, it states: $A^x + B^y = C^z$, where $A, B, C, x, y$ and $z$ are positive integers and $x, y$ and $z$ are all greater than $2$, then $A, B$ and $C$ must have a common prime factor.

The following will lead to my question:

Let us have a consecutive set $A$ of elements $x^2$: $1, 4, 9, 16, 25, 36 , 49, 64 ,81, 100…$

The set’s elements are growing in a rate, that results in the following set $B$ with elements $y$ : $3, 5 , 7,$ *9 $, 11, , 13 , 15, 17, 19$ (growing at $y+2$)

*Note $9 = 3^2$

Since the Beal’s conjecture has no restrictions on $x^2$

Let’s move to a consecutive set $A$ of elements $x^3$: $1, 8, 27, 64, 125, 216, 343 , 512 ,729, 1000…$

The set’s elements are growing in a rate that results in the following set $B$ with elements $y$: $7, 19 , 37, 61, 91, 127 ,$ *169 $, 217, 271 …$ (growing $12, 18, 24, 30 …$)

Needless to demonstrate other sets of $x^z$: is there a way to prove that $y$ in set $B$ can only result as $x^2$ such as *$169$ which is $13^2$ and cannot be any other $x^z$ where $z$ is bigger than $2$ (such as: $x^3$, $x^4$ ….)?

It is way above my level to try and prove the Beal’s conjecture and I am not trying to demonstrate any general proof, however my question is whether it can be proved that at least for any

$x^z – (x-1)^z$, it can only end in $x^2$ and never in $x^z$ where $z$ > 2?

If not, whether it can be partially proved for a specific $z$, example such as in the case of $x^3 – (x-1)^3$ where $z=3$?