TITLE: An equation involving perfect numbers
QUESTION [5 upvotes]: Let $s,x_1,x_2,\cdots, x_s$ be natural numbers not neccesarily distinct.
I am interested in solving the equation $$(x_1+x_2+\cdots +x_s)^s=2^s(x_1\cdot x_2\cdots x_s)^2$$
Some Notes: 
I have found two solutions $(x_1,\cdots ,x_s)$
1) We can see that  equality is satisfied  if $x_s= 2^{p-1}(2^p-1)$ is an even perfect number and all the other $x_i$ are the  proper divisors of $x_s$. (That is why i started to investigate the equation)
Of course we can disregard the restriction that $x_s$ is perfect and show easily that we also get a solution if:   $$x_1=1,x_2=2^1,...,x_n=2^{n-1},x_{n+1}=2^n-1,x_{n+2}=2^1(2^n-1),...$$
and  $x_s=2^{n-1}(2^n-1)$.(The number $n$ is not necessarily prime )     
2)  A  trivial one: $$x_1=x_2=\cdots =x_s=\frac{s}{2}$$ .   
Is it possible to find other solutions or to prove that there are only  2 solutions, those mentioned above?
Thanks in advance.

REPLY [8 votes]: I think the computer found other solutions for $s=4$:
[x1,x2,x3,x4]
[ 2 , 27 , 150 , 1 ]
[ 2 , 3 , 6 , 121 ]
[ 27 , 150 , 2 , 1 ]
[ 50 , 6 , 3 , 1 ]