Find all Pairs of Integers $(x,y)$ , $x \gt y \ge 2$ such that $x^y=y^{x-y}$ Find all Pairs of Integers $(x,y)$ , $x \gt y \ge 2$ such that $x^y=y^{x-y}$
My Try: 
if $y=2$ then $$x^2=2^{x-2}$$ and Taking log on both sides we get
$$2 \log x=(x-2) \log 2$$ i.e.,
$$2 \log_2 x=x-2$$ if $x=2^k$ then we get
$$2k=2^k-2$$ or
$$k+1=2^{k-1}$$
By Trying different values we get $k=3$ is the only solution.
so one pair is $(x,y)=(8,2)$
Similarly if $y=3$ we get
$$k+1=3^{k-1}$$ which is satisfied by $k=2$ so another pair is
$(x,y)=(9,3)$
Is this approach correct?

For
$x^y=y^{x-y}
$
I find that
$x=8, y=2$
and
$x=9, y=3$
are the only solutions.
Must have $x > y$.
$x^y
=y^{x-y}
$
so
$(xy)^y = y^x
$.
Let
$x = ry$ where
$r > 1$.
$(ry^2)^y = y^{ry}$
so
$ry^2 = y^r$
or
$r = y^{r-2}$
or
$y = r^{1/(r-2)}$.
If $x = 2y$
then
$(2y)^y = y^y$
which has no solution.
If
$1 < r < 2$,
then
$r-2 < 0$
so
$y = r^{1/(r-2)}
< 1$,
so no solution.
Therefore $r > 2$.
$r^{1/(r-2)} = 2$
for $r = 4$.
Since
$r^{1/(r-2)} $
is decreasing for
$r \ge 4$,
must have
$2 < r \le 4$.
If $r=4$
then $y=2$
and $x = ry = 8$.
$(xy)^y = (16)^2 = 256$
and
$y^x = 2^8 = 256$
so this is a solution!
If
$r^{1/(r-2)} = 3$,
then
$r=3$.
Then
$y = 3$
and $x = ry = 9$.
$(xy)^y = 27^3 = 3^9$
and
$y^x = 3^9$,
so this is a solution!
Let
$d = (x, y)$
so
$x = ad, y = bd$
with
$4b >a > 2b, (a, b) = 1$.
From
$(xy)^y = y^x
$,
$(abd^2)^{bd} = (bd)^{ad}
$
or
$(abd^2)^{b} = (bd)^{a}
$
or
$a^bb^bd^{2b} = b^ad^a$
or
$a^b = b^{a-b}d^{a-2b}$.
Must have
$b=1$
since
$(a, b) = 1$.
This becomes
$a = d^{a-2}$
or
$d = a^{1/(a-2)}$.
As before,
the only integer solutions
to $d = a^{1/(a-2)}$
are
$a=4, d=2$
and
$a=3, d=3$
which we have found already.
Therefore 
$x=8, y=2$
and
$x=9, y=3$
are the only solutions.
Note on the
derivative of
$r^{1/(r-2)}$.
$\begin{array}\\
(r^{1/(r-2)})' 
&= r^{1/(r-2)} (\frac1{r(r-2)}-\frac{\log(r)}{(r-2)^2}\\
&= r^{1/(r-2)} \frac{(r-2)-r\log(r)}{r(r-2)^2}\\
& < 0
\qquad\text{for } r > 2\\
\end{array}
$
so
$r^{1/(r-2)} < 2$
for
$r > 4$.
