TITLE: Integer Solutions of $x+y^n = y + x^m$ for $n < m$
QUESTION [13 upvotes]: I found 8 of them and believe there is no more:
$$2+3^2=3+2^3$$
$$2+6^2=6+2^5$$
$$6+15^2=15+6^3$$
$$3+16^2=16+3^5$$
$$3+13^3=13+3^7$$
$$2+91^2=91+2^{13}$$
$$5+280^2=280+5^7$$
$$30+4930^2=4930+30^5$$
I call the solution a principal pipe of order 2. Any idea to attack this equation?
If $n=x=2$, it becomes $2+y^2=y+2^m$. It is Ramanujan and Nagell Equation, and there are only 3 solutions. If $n=2$ and $x \geq 3$, then we can use the similar way to prove that there is no other solution for $x < 10000$ with help of computer. 
If $n=3$, and we assume $x$ and $y$ are prime, and $6\mid (m-1)$, then the solution is $3+13^3=13+3^7 = 2200$. 
I believe there is no solution for $n>3$. I am particularly interested in solutions that both $x$ and $y$ are odd primes.

REPLY [11 votes]: This problem was considered in a paper of Mignotte and Petho {Publ. Math. Debrecen 1999) and subsequently in one of mine [Canad. Math. J 2001], where there is a conjecture that the equation $a^x-b^y=c$ has, for fixed positive integers $a, b$ and $c$, with $a, b \geq 2$, at most one solution in positive integer exponents $x$ and $y$, with precisely $11$ exceptions. This is provable if $c$ is suitably large or suitably small, but, as far as I know, is still open in general. This problem arose from four classical papers of Pillai.