This is not an answer to your question but at least a reference to another thread where this problem has also been discussed: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=620389

The positive solutions of the Diophantine equation $(1) \;\; x^y y^x = (x + y)^z$ are given by $(x,y,z) = (2^{2^n - 1}, 2^{2^n - 1}, 2^{2^n - n}(2^n - 1))$, where $n$ is a positive integer. Proof: Let $d=GCD(x,y)$. Hence there exist two coprime positive integers $r$ and $s$ s.t. $x=rd$ and $y=sd$, which according to (1) means $(rd)^y (sd)^x = [d(r+s)]^z$, i.e. $(2) \;\; d^{x+y-z} r^y s^x = (r + s)^z$. Observe that $GCD(rs,r+s)=1$ since $GCD(r,s)=1$. Let us consider the following two cases: Case 1: $x+y-z < 0$. Then by (2) $(3) \;\; d^{z-x-y} = \frac{r^y s^x}{(r+s)^z} \in \mathbb{N}$ since $z-x-y >0$. Combining (3) with the fact $GCD(rs,r+s)=1$ yields $(r+s)^z=1$, which is impossible since $r,s,z\geq 1$. Hence there is no solution of (1) in this case. Case 2: $x+y-z \geq 0$. Then by (2) $(4) \;\; d^{x+y-z} = \frac{(r+s)^z}{r^y s^x} \in \mathbb{N}$. Combining (4) with the fact $GCD(rs,r+s)=1$ yields $r^y s^x = 1$, implying $r=s=1$. Consequently $x=y=d$, and thus by (4) $(5) \;\; d^{2d-z} = 2^z$. Clearly $d=2^m$ for a positive integer $m$, which inserted in (5) result in $2^{m(2^{m+1} - z)} = 2^z$. Hence $m(2^{m+1} - z) = z$, yielding $(6) \;\; z = \frac{2^{m+1}m}{m+1}$. Now $GCD(m,m+1)=1$, which according to (6) means ${\textstyle \frac{2^{m+1}}{m+1} \in \mathbb{N}}$. Therefore $m+1 = 2^n$ for a positive integer $n$, which inserted in (6) give us $z = 2^{2^n - n}(2^n - 1)$, where $x = y = d = 2^m = 2^{2^n - 1}$. $\;\;$ q.e.d.

LHS is divisible by x, RHS is also divisible by x, so y is divsible by x. LHS is divisible by y, RHS is also divisible by y, so x is divsible by y. so x=y. is that right?

sebepkaly wrote: LHS is divisible by x, RHS is also divisible by x, so y is divsible by x. LHS is divisible by y, RHS is also divisible by y, so x is divsible by y. so x=y. is that right? No

Here is my solution: https://beyondmathsolutions.wordpress.com/2015/02/17/soln-2015-kazakhstan-nmo-problem-2/ (how do i include hyperlinks in a post?)