Find all $x,y\in \mathbb{N}$, such that $x^{y}=y^{y-x}$

Find all $x,y\in \mathbb{N}$, such that $ x^{y}=y^{y-x} $.

I have found the solutions $(1,1)$ and $(2,4)$. I've been trying to prove that there aren't any more. I tried splitting the cases by parity, but nothing interesting seemed to come out.

I also tried rearranging it into:

$x=y^{\frac{y-x}{y}}$.

Is there any way to proceed with this problem?

Thank you very much

Actually, the function can be written as $$ \left(\frac{y}{x}\right)^{y} = y^{x} $$ If you let $t = \frac{y}{x}$ then we can get $$ t^{t} = y $$ So if you apply $y = t^{t}$ and $x = t^{t - 1}$ with $t$ an integer you get infinite pairs of solutions. Furthermore, if $t$ is not an integer then $t = q/p$ which is irreducible. Then $y = (q/p)^{q/p}$ is an integer so $y^{p} = (q/p)^q$ is an integer, which is not possible. So $(x, y)$ is a solution to the equation iff $(x, y) = (t^{t-1}, t^t)$ with $t$ an integer, and there are infinitely many pairs.