Do there exist an infinite number of integer-solutions $(x,y,z)$ of $x^x\cdot y^y=z^z$ where $1\lt x\le y$? Question : Do there exist an infinite number of integer-solutions $(x,y,z)$ of $x^x\cdot y^y=z^z$ where $1\lt x\le y$ ?
Motivation : After struggling to find a solution, I've just got one solution, which is 
$$(x,y,z)=(1679616, 2985984, 4478976).$$
In the following, I'm going to write how I got this solution.
Letting $d$ be the greatest common divisor of $x,y,z$, we can represent
$$x=ad, y=bd, z=cd$$
where $a,b,c$ are coprimes with each other. Then, we get
$$d^{a+b-c}\cdot a^a\cdot b^b=c^c.$$
In the following, let's consider without the condition $x\le y$. Here, I suppose
$$a=2^m, b=3^n, a+b-c=1.$$
(As a result, this supposition works.)
Then, we get
$$d=\frac{c^c}{2^{ma}\cdot 3^{nb}}.$$
Hence, letting $c=2^k\cdot 3^l$, if 
$$kc\ge ma=m\cdot 2^m, lc\ge nb=n\cdot 3^n,$$
then $d$ is an integer.
Since $(m,n)=(4,2)$ satisfies the above conditions, then we get $d=2^8\cdot 3^6=186624.$ Hence we can get 
$$x=9d=2^8\cdot 3^8=1679616, y=16d=2^{12}\cdot 3^6=2985984, z=24d=2^{11}\cdot 3^7=4478976.$$
Note that here I interchanged $x$ and $y$.
P.S : I was surprised to get this solution because I got this almost by chance. So, I don't know the other solutions. If you have any helpful information, please teach me.
 A: Ke Zhao find these solutions: $n\in\mathbb N,$
$$x=2^{2^{n+1}(2^n-n-1)+2n}(2^n-1)^{2(2^n-1)},\\
y=2^{2^{n+1}(2^n-n-1)}(2^n-1)^{2(2^n-1)+2},\\
z=2^{2^{n+1}(2^n-n-1)+n+1}(2^n-1)^{2(2^n-1)+1}.$$
You may find that these solutions satisfies $4xy=z^2,$ W.H.Mills proved that 
1)If $4xy=z^2$ then Ke Zhao's solutions are all of them.
2)If $4xy>z^2$ then there has no solutions.
3)If $4xy<z^2$ then there only exist fintite many solutions.
Hence maybe Ke Zhao has given all the solutions to $x^xy^y=z^z,$ but this has not been proven.
A: I've just got the following in a similar way as above.
The answer is Yes; there exist an infinite number of solutions.
Letting $p,q$ be coprime integers with each other, suppose 
$$a=p^m, b=q^n, a+b-c=1, c=p^k\cdot q^l, kc\ge ma, lc\ge nb.$$
Here, I got the following because of a relational equation $N^2+(N-1)^2=2N(N-1)+1$.
$$p=N-1, m=2, N=2^s, q=2, n=2s, k=1, l=s+1,$$$$ a=(2^s-1)^2, b=2^{2s}, c=2^{s+1}(2^s-1).$$
Now, we know that if $s\ge 2$, then these satisfy the above conditions. Also, note that the $s=2$ case is the case I found.
Hence, we get
$$d=2^u\cdot p^v, u=2^{s+1}(2^s-s-1), p=2^s-1, v=2(2^s-1),$$$$ x=2^u\cdot p^{v+1}, y=2^{2s+u}\cdot p^v, z=2^{u+s+1}\cdot p^{v+1}.$$
You can take every integer larger than or equal to $2$ as $s$, then we now know that there exist an infinite number of solutions. Now the proof is completed.
$P.S.$ For $s=3$, we get
$$p=7, a=49, b=64, c=112, u=64, v=14,$$$$ x=2^{64}\cdot 7^{16}, y=2^{70}\cdot 7^{14}, z=2^{68}\cdot 7^{15}.$$
Note that $z=2^{68}\cdot 7^{15}\approx 10^{33}.$
I'm interested in finding every solution, which seems very hard.
