In this post we denote the Dedekind psi function as $\psi(m)$ for integers $m\geq 1$. This is an important arithmetic fuction in several subjects of mathematics. As reference I add the Wikipedia Dedekind psi function, and [1].
The Dedekind psi function can be represented for a positive integer $m>1$ as $$\psi(m)=m\prod_{\substack{p\mid m\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$$ with the definition $\psi(1)=1$.
Claim. If we take $x=2k$ and $y=2x$ for a given integer $k\geq 1$, then the equation
$$y^{y-x}(y-x)^y=\frac{y-x}{\psi(y-x)}\cdot\psi(x^y y^x),\tag{1}$$ holds.
Sketch of proof. Just direct computation using the mentioned representation for the Dedekind psi function.$\square$
Question. I would like to know a characterization of the solutions of the equation $(1)$ for integers $x$ and $y$ satisfying $1\leq x<y$. In other words, I would like to know if the solutions of $(1)$ are of the form $(x,y)=(2k,4k)$ where $k\geq 1$ runs over positive integers. Are these all the solutions? Can you find a solution $(x,y)$, for integers $1\leq x<y$, of a different form? Many thanks.
With a Pari/GP script and for small segments of integers I have not found examples of other solutions different than $(2k,4k)$ for integers $k\geq 1$.
References:
[1] Tom M. Apostol, Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag (1976).
This is a partial answer.
This answer proves the following claims :
Claim 1 : If $y=2x$ with odd $x$, then $(1)$ does not hold.
Claim 2 : If $y\not=2x$ and $y\not=2x-1$, then $y$ is a composite number.
Claim 3 : If $y\lt 3x-1$ and $y\not=2x$, then $x$ is a composite number.
Let us use the following lemma :
Lemma : If $n\ge 2$, then $P(\psi(n))\le n+1$. If $n=3$ or $n\ge 5$, then $P(\psi(n))\le \frac{n+1}{2}$ where $P(n)$ is the largest prime factor of $n$.
Proof for lemma : If $n=2,4$, then $P(\psi(n))=3\le n+1$.
If $n=2^k$ with $k\ge 3$, then $$P(\psi(n))=P(\psi(2^k))=P(3\cdot 2^{k-1})=3\le\frac{n+1}{2}$$
If $n=p$ where $p$ is an odd prime, then $$P(\psi(n))=P(\psi(p))=P(p+1)\le \frac{p+1}{2}=\frac{n+1}{2}$$
If $n=p^k$ where $p$ is an odd prime with $k\ge 2$, then $$P(\psi(n))=P(\psi(p^k))=P(p^{k-1}(p+1))=p\le\frac n3\le\frac{n+1}{2}$$
If $\displaystyle n=\prod_{i=1}^{d}p_i^{e_i}$ where $d\ge 2$ and $p_1\lt p_2\lt \cdots\lt p_d$ are primes, then $$P(\psi(n))=P\bigg(\psi\bigg(\prod_{i=1}^{d}p_i^{e_i}\bigg)\bigg)=P\bigg(\prod_{i=1}^{d}p_i^{e_i-1}(p_i+1)\bigg)\le p_d\le\frac n2\le\frac{n+1}{2}$$ since $p_d+1$ is an even composite number. $\quad\blacksquare$
Claim 1 : If $y=2x$ with odd $x$, then $(1)$ does not hold.
Proof : $(x,y)=(1,2)$ is not a solution. If $x$ is odd larger than $1$, then we get $2\psi(x)=3\psi(x)$ which does not hold. $\quad\blacksquare$
Claim 2 : If $y\not=2x$ and $y\not=2x-1$, then $y$ is a composite number.
Proof : Since $\psi(x^y y^x)=x^{y-1}y^{x-1}\psi(xy)$, the equation $(1)$ is equivalent to
$$y^{y-x}(y-x)^{y-1}\psi(y-x)=x^{y-1}y^{x-1}\psi(xy)$$
Suppose that $y$ is a prime number.
Case 1 : $y\le 2x-2$. Then, we have$$(y-x)^{y-1}\psi(y-x)=x^{y-1}y^{2x-y-1}\psi(xy)$$So, $\psi(y-x)$ has to be divisible by $y$. If $y-x=1$, then since $x\ge 3$, we see that $1=x^{x}(x+1)^{x-2}\psi(x(x+1))$ has no solutions. It follows from $y-x\ge 2$ and Lemma that $P(\psi(y-x))\le y-x+1$. If $x\ge 2$, then $P(\psi(y-x))\le y-1$ from which $y\not\mid \psi(y-x)$ follows. So, we get $x=1$ and $y\le 0$ which is impossible.
Case 2 : $y\gt 2x$. Then, we have $$y^{y-2x+1}(y-x)^{y-1}\psi(y-x)=x^{y-1}\psi(x)(y+1)$$ So, $\psi(x)$ has to be divisible by $y$. If $x=1$, then $y^{y-1}(y-1)^{y-1}\psi(y-1)=y+1$ which has no solutions since for $y\ge 3$, then LHS is larger than RHS. It follows from $x\ge 2$ and Lemma that $P(\psi(x))\le x+1$. If $y-x\ge 2$, then $P(\psi(x))\le x+1\le y-1$ from which $y\not\mid\psi(x)$ follows. So, we have to have $y-x=1$ and $x\lt 1$ which is impossible.
From the two cases above, the claim follows. $\quad\blacksquare$
Claim 3 : If $y\lt 3x-1$ and $y\not=2x$, then $x$ is a composite number.
Proof : Suppose that $x$ is a prime number.
Case 1 : $y\le 2x-2$. Then, we have$$(y-x)^{y-1}\psi(y-x)=x^{y-1}y^{2x-y-1}\psi(xy)$$ If $y-x=1$, then $1=x^{x}(x+1)^{x-2}\psi(x)\psi(x+1)$ which is impossible. Since $y-x\ge 2$, we have $y-x\lt x$ and $P(\psi(y-x))\le y-x+1\lt x$, and so LHS is not divisible by $x$.
Case 2 : If $y=2x-1$, then $$(x-1)^{2x-2}\psi(x-1)=x^{2x-2}\psi(x)\psi(2x-1)$$ So, $\psi(x-1)$ has to be divisible by $x$. From Lemma, if $x-1=3$ or $x-1\ge 5$, then $P(\psi(x-1))\le\frac{x}{2}\lt x$, so we have to have $x-1=1,2,4$ for which the equation does not hold.
Case 3 : $2x\lt y\lt 3x-1$. Then, we have $$y^{y-2x+1}(y-x)^{y-1}\psi(y-x)=x^{y-1}\psi(x)(y+1)$$ If $y-x\not=1,2,4$, then $P(\psi(y-x))\le\frac{y-x+1}{2}\lt x$. So, since $x$ is prime, we have to have $y=kx$ to have $2\lt k\lt 3-\frac 1x$ which is impossible. If $y-x=1$, then $(x+1)^{2-x}=x^{x}\psi(x)(x+2)$ has no solutions $x\le 2$. If $y-x=2$, then $(x+2)^{3-x}\cdot 2^{x+1}\cdot 3=x^{x+1}\psi(x)(x+3)$ has no solutions $x\le 3$. If $y-x=4$, then $(x+4)^{5-x}\cdot 2^{2x+6}\cdot 6=x^{x+3}\psi(x)(x+5)$ has no solutions $x\le 5$.
From the three cases above, the claim follows. $\quad\blacksquare$