Solve the equation $x^{x+y}=y^{y-x}$ over natural numbers 
Solve the equation $x^{x+y}=y^{y-x} \tag 1$ where $x,y \in \mathbb{N}$


$x = 1,y = 1$ is a solution, now suppose $x \ne 1, y \ne 1$.
Obviously $x, y \ne 0$ and $x, y$ have same prime divisors. 
Because $x+y \gt y-x$ it follows that $x \mid y$ therefore $y = kx$ and (1) becomes: $x^{(k+1)x}=(kx)^{(k-1)x} \tag2$ and, dividing by $x^{k-1}$ we get
$x^{2x}=k^{(k-1)x} \tag 3$ therefore $x^2=k^{k-1} \tag 4$ and here I've got stuck. Any help is appreciated.
UPDATE
From (4) we have $k \ge 3$ therefore $2 \le k-1$ and $ k \mid x$. Let $x=km$ and (4) becomes $k^2m^2=k^{k-1} \tag5$ and $m^2=k^{k-3} \tag6$
UPDATE 2
$x=t^{t^2-1}, y=t^{t^2+1}$ is solution $\forall t \in \mathbb{N}-\{0\}$

From
$x^2=k^{k-1}$,
$k$ must be odd.
Let $k = 2j+1$,
so
$x^2 = (2j+1)^{2j}$
or
$x = (2j+1)^j$.
Then
$y = kx
=(2j+1)(2j+1)^j
= (2j+1)^{j+1}
$.
Check:
$x^{x+y}?y^{y-x}$
$\begin{array}\\
x^{x+y}
&=((2j+1)^j)^{(2j+1)^j+(2j+1)^{j+1}}\\
&=(2j+1)^{j(2j+1)^j(1+(2j+1))}\\
&=(2j+1)^{j(2j+1)^j(2j+2)}\\
&=(2j+1)^{2j(j+1)(2j+1)^j}\\
\end{array}
$
$\begin{array}\\
y^{y-x}
&=((2j+1)^{j+1})^{(2j+1)^{j+1}-(2j+1)^{j}}\\
&=(2j+1)^{(j+1)((2j+1)^{j+1}-(2j+1)^{j})}\\
&=(2j+1)^{(j+1)(2j+1)^{j}((2j+1)-1)}\\
&=(2j+1)^{(j+1)(2j+1)^{j}(2j)}\\
&=(2j+1)^{2j(j+1)(2j+1)^{j}}\\
\end{array}
$
And they match!
