Existence of Pillai equations with Catalan type solutions? In Catalan's conjecture we have $$x^m-y^n=1$$ having solution $(3,2,1,1)$ and $(3,2,2,3)$.
Call $$ax^m-by^n=k$$ to be Pillai Diophantine equation.


*

*Is it true no Pillai Diophantine equation exists with integer solutions $(x,y,m,n)$ and $(x,y,m+1,n+r)$ with $1<r$ true at $a=b=1\neq k$?


*Is it also true at any fixed $a,b\in\Bbb Z$ and at any fixed $k\in\Bbb Z$ with $abk\neq0$ or $|a-b|\neq |k|$?

Following Jeremy Rouse's post both are no. His answer also works $k>0$ for 2. if $a=-2$ and $b=-7$ which has $b<a$.
I suspect there is no further equations with $k>0$ and $a=b=1$ or $k>0$ with $b\geq a$. I assume this to be true (without proof).
However his solution has $(m+1,n+r)-(m,n)=(1,2)$ and the main query I seek is whether we can get $(m+1,n+r)-(m,n)\neq (1,2)$ and $gcd(m,n)=gcd(m+1,n+r)=1$ at:



*$a=b=1$ and $k>0$?


*any other $a,b$ with $a,b,k>0$?

 A: The answer to questions 1 and 2 are both no. In the example you consider, where $m = 1$, $n=1$ and $r = 2$, you are seeking solutions to $x-y = x^{2} - y^{3} = k$. The equation $x-y = x^{2} - y^{3}$ is an elliptic curve and the largest integral point on this curve gives you a solution for $k = -20$, namely
$$
  -20 = (-14) - 6 = (-14)^2 - 6^3.
$$
Variants of this trick work for other values of $a$ and $b$. For example, if $a = 2$ and $b = 7$ we find that
$$ -86968 = 2 \cdot (-40754) - 7 \cdot 780 = 2 \cdot (-40754)^{2} - 7 \cdot 780^{3}. $$
Indeed, there are $9$ different values of $k$ that work for $a = 2$, $b = 7$, $m = 1$, $n = 1$ and $r = 2$.
EDIT: The answers to questions 3 and 4 are also both no.
For one thing, the "trivial solution" with $x = 1$ and $y = -1$ gives $k = 2$ whenever $n$ and $n+r$ are both odd. However, there are many non-trivial solutions too.
For example, if $m = n = 1$ and $r = 4$ with $a = b = 1$, we have solutions with $k = 4$ and $k = 13$, namely
$$ 4 = 6-2 = 6^{2} - 2^{5}, \quad 13 = 16 - 3 = 16^{2} - 3^{5}. $$
This shows the answer to question 3 is no.
If we take $a = 2$ and $b = 7$, with $m = n = 1$ and $r = 4$ we find
$$ 175 = 2 \cdot 105 - 7 \cdot 5 = 2 \cdot 105^{2} - 7 \cdot 5^{5}. $$
This shows the answer to question 4 is no too.
