Diophantine equation-problem 1

I do not have a full proof. Basically, if $x \neq y$ , then unless both variables are quite 'small,' one of the first two terms (whichever has the greater exponent) dominates all the other terms and quickly prevents equality. So the search range for positive integer solutions can be restricted to a small set (eg. both $x$ and $y$ must be less than $10$ ), and then checked by brute force.

Consider the case of $y > x \geq 3$ : $3^4 - 4^3 = 17$ , $3^5 - 5^3 = 118$ , $4^5 - 5^4 = 399$ , ... Once this difference becomes greater than $y^3 > xy^2 > xy^2 - 19$ , there is no need to check for further solutions. It is pretty well known that $x^y > y^x$ for $y > x$ once both variables are greater than $e$ , so greater than $3$ in the case of integers. Because of the terms on the right hand side, you have to check a bit farther than $3$ . I think it would be sufficient to check until $x^y > 2y^x$ for $y > x$ . This is where I need some help.

The cases of $x = 1,2$ or $y=1,2$ can be handled separately, which end up leading to some solutions.