Find the positive integer solutions for the equation $x^y-y^x=1$

Question

Find all positive integer $x,y$ such $$x^y-y^x=1$$

It is clear $(x,y)=(2,1)$ and $(x,y)=(3,2)$ hold.

Answer 1

We know that for any two positive integers $x,y$ if $x=y$ then $x^y-y^x=0$, hence possible solutions of $x^y-y^x=1$, for $x>3$ should be expressible in form $y=x+k$, where $k-$positive integer. $$ x^{x+k}-(x+k)^{x}-1=0 $$ Motivation behind setting $k$ to be positive integer is that for any integer $x>3$ expression above strictly lesser than $0$. Also, notice that for $x>3$ addendum with higher power is growing faster, namely if $k<0$ whole expression tends to $-\infty$ and if $k>0$ it tends to $\infty$ as $x$ is growing. Thus, what we'd like to show is $x^{x+1}-(x+1)^x-1>0$, for any $x>3$, if it's true then no solutions exist. Once again, rate of growth is higher for $x^{x+1}$ and in case of $x=4$: $4^5-5^4-1=498>0$.

Hence, there is no solutions except mentioned $(2,1)$ and $(3,2)$.