On a symmetric equation over the integer lattice that involves the Euler's totient function I would like to know hints or a proof, or counterexamples, for the conjecture that I've stated in the Question below. I'm interested in this in an attempt to continue the study of a question that I've posted on MathOverflow (the corresponding to the link for the cited thread of comments in the following paragraphs). The first I add the following claim, since our conjecture is the converse of this.
Claim. If $(x,y)$ is a point of the integer lattice (I mean that both $x$ and $y$ are integers) with $x\geq 1$ an odd integer and $x<y$, then the equation $$y^{y-x}(y-x)^y=2\cdot\frac{y-x}{\varphi(y-x)}\,\varphi(x^y y^x)\tag{1}$$
holds for $y=2\cdot x$, where $\varphi(m)$ denotes the Euler's totient function.
Sketch of proof. I'm inspired as starting point in the thread of comments$^{*}$ that I've posted in my question with identificator 336501 on MathOverflow and title What work can be done to study the solutions of $\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)$? (asked Jul 19 '19). From here it is easy to check, for lattice points of the form $(x,y=2x)$ with $x$ an odd integer, that $$y^xx^y=y^x\left(\frac{y}{2}\right)^y=2\cdot\frac{y-x}{\varphi(y-x)}\varphi(x^y y^x)$$
holds since $y=2x$, and from here we get $(1)$ because $y-x=x$.$\square$

Question. I would like to know what work can be done to state the veracity of the following conjecture (converse of previous claim): If $(x,y)$ is a point of the integer lattice, I mean that both $x\geq 1$ and $y\geq 1$ are integers, with $x<y$, and the equation $$y^{y-x}(y-x)^y=2\cdot\frac{y-x}{\varphi(y-x)}\,\varphi(x^y y^x)$$
holds, then $x$ is an odd integer and $y$ is twice times $x$ (in simple words that our points are of the form $(x,y=2\cdot x)$ for $x\geq 1$ and odd integer). Many thanks.

If my question is feasible but tedious to get, feel free to add just a summary of hints, or a key idea. In other case add what work can be done about the veracity of the conjecture.
Remarks and computational evidence. The motivation to ask such thing (I'm agree that this seems an artificious characterization of those points of the integer lattice having the form $(\text{odd integers},\text{twice the odd numbers})$) is my study of the mentioned MO post. I can to test the conjecture that I've asked in Question for the segment of integers $1\leq x<y\leq 1000$ .

$^*$Thus, as aside remark (unrelated to our post), on assumption that there exists an odd perfect number $X$, the equation $$X^Y Y^X=(Y-X)^Y Y^{Y-X}=\frac{\sigma(X)}{\varphi(Y-X)}\,\varphi(X^Y Y^X)$$
holds for the lattice point $(X,Y=2X=\sigma(X))$, where $\sigma(m)=\sum_{1\leq d\mid m}d$ denotes the sum of positive  divisors function of an integer $m\geq 1$.
 A: The conjecture is true.
Proof : 
Firstly, let us prove that $2x-y\ge 0$.
Suppose that $2x-y\lt 0$.
If $y-x=2$, then $(x,y)=(1,3)$ for which the equation does not hold.
If $y-x\ge 3$, then since $\varphi(y-x)\ge 2$, we get
$$\begin{align}& y^{y-x}(y-x)^{y-1}\varphi(y-x)=2\varphi(x^y y^x)
\\\\&\implies 2y^{y-x}(y-x)^{y-1}\le y^{y-x}(y-x)^{y-1}\varphi(y-x)=2\varphi(x^y y^x)\lt 2x^yy^x
\\\\&\implies 2y^{y-x}(y-x)^{y-1}\lt 2x^yy^x
\\\\&\implies y^{y-2x}\lt x\bigg(\frac{x}{y-x}\bigg)^{y-1}
\\\\&\implies y\le y^{y-2x}\lt x\bigg(\frac{x}{y-x}\bigg)^{y-1}\lt x
\\\\&\implies y\lt x\end{align}$$
which is impossible. So, we have to have $2x-y\ge 0$.$\quad\square$
Secondly, let us prove that $2x-y\le 1$.
Suppose that $2x-y\gt 1$.
Using the fact that
$$\varphi(x^y y^x)=x^{y-1} y^{x-1}\varphi(xy)$$
we see that the equation can be written as
$$(y-x)^{y-1}\varphi(y-x)=2x^{y-1}y^{2x-y-1}\varphi(xy)\tag1$$
Since $\varphi(y-x)\lt y-x\lt x\le x^{y-1}$, we have $x^{y-1}\not\mid \varphi(y-x)$. So, we have to have $x^{y-1}\mid (y-x)^{y-1}$. It follows that there is a positive integer $k$ such that $y=kx$. It follows from $2x-kx\gt 1$ that $k=1$ for which LHS of $(1)$ equals $0$ while RHS of $(1)$ is positive. So, we have to have $2x-y\le 1$.$\quad\square$
Thirdly, let us prove that $2x-y\not=1$.
Suppose that $2x-y=1$. Then, we get
$$(x-1)^{2x-2}\varphi(x-1)=2x^{2x-2}\varphi(x(2x-1))$$
If $x=1$, this does not hold. If $x\gt 1$, we have to have
$$x^{2x-2}\mid \varphi(x-1)\implies x^{2x-2}\le \varphi(x-1)\lt x$$
which is impossible. So, we have to have $2x-y\not=1$.$\quad\square$
It follows that $y=2x$.
We can write $x=2^ab$ where $a$ is a non-negative integer and $b$ is a positive odd integer.
Suppose that $a\ge 1$. Then, we get
$$(2x)^{x}(x)^{2x-1}\varphi(x)=2\varphi(x^{2x} (2x)^x)\implies 2^{3ax+x-1}= 
 2^{3ax+x}$$
which is impossible. So, we have to have $a=0$.
Therefore, $x$ has to be odd with $y=2x$.$\quad\blacksquare$
