Find integers $x > 1, y > 1$ such that $(y^x+1) \mid (x^y+1)$

Question

A problem from quora.

Find integers $x > 1, y > 1$ such that $(y^x+1) \mid (x^y+1)$.

I have found some infinite classes of solutions, but none with $x$ and $y$ relatively prime.

Here are the ones I have found so far. I know how I could construct more, but the algebra gets annoying.

$$(x, y) =(2n+1, (2n+1)^{2n+1}),\\ (2(2n+1), (2(2n+1))^2),\\ (3(2m+1), (3(2m+1))^3). $$

Answer 1

\begin{align*}\frac{x^y + 1}{y^x + 1}&=k\\\\ \text{Let}\qquad (x^y+1)&=ky^x+k \quad x,y,k>0\\ x^y-ky^x&=k-1 \end{align*}

We let

$\quad k=1\implies x^y\,-\,\,y^x=0\implies x=y$ $\quad k=2\implies x^y-2y^x=1\implies x=3,y=1$ $\quad k=3\implies x^y-3y^x=2\implies x=5,y=1$ $\quad k=4\implies x^y-4y^x=3\implies x=7,y=1$

We can see a pattern in this near trivial solution. If $\,y=1\,$ the denominator will be $\,2\,$ and any odd number $\,x\,$ will yield a solution.

Answer 2

We want to find integer solution to $$\quad \frac{x^y + 1}{y^x + 1}$$ Here are some solutions

$$(2,2)\quad (4,2)\quad (3,3)\quad (2,4)\quad (4,4)\quad (5\le x\le 26,x)\\ (\{3,49,18,27\},27)\qquad (\{11,14,28\},28)\qquad (\{7,8,9,10,11,29\},29)\\ (\{6,7,8,9,10,11,12,13,29\},30)\qquad (\{5,6,7,8,9,10,11,12,13,14,15,31\},31)\\ (\{4,5,6,7,8,9,10,11,12,13,14,15,16,32\},32)\quad $$ I cannot see a pattern but they are easy to find in a spreadsheet.