Equation with Vieta Jumping: $(x+y+z)^2=nxyz$. 
Find all $n\in\mathbb{N}$ so that there exist $x,y,z\in \mathbb{N}$ that solve:
  $$(x+y+z)^2=nxyz$$

I tried to attack it finding solutions, but all the solutions doesn't seem to have something in common. For example:
$$ (x,y,z,n)=(1,1,1,9)$$
$$ (x,y,z,n)=(1,2,3,6)$$
$$ (x,y,z,n)=(1,4,5,5)$$
$$ (x,y,z,n)=(2,2,4,4)$$
$$ (x,y,z,n)=(8,8,16,1)$$
I was trying to solve this problem, but someone told me that a hint to solve it is using Vieta Jumping. I don't know how to use Vieta Jumping, so can anyone help me to understand how to apply Vieta Jumping and solve the problem?

Fair to say this one is difficult; a full proof would need everything in HURWITZ 1907. 
Each fundamental solution is given as $x \geq y \geq z \geq 1,$ with
$$  nyz - 2y- 2z \geq 2x, $$ or
$$ 2(x+y+z) \leq nyz.  $$ Each fundamental solution is the base of an infinite tree of solutions; given $(n;x,y,z),$ we get three solutions
$$  (n; \; \; nyz - 2y- 2z-x, \; \; y, \; \; z), $$
$$  (n; \; \; x, \; \;  nzx - 2z- 2x-y, \; \; z), $$
$$  (n; \; \; x, \; \; y, \; \; nxy - 2x- 2y-z). $$
The trees are quite similar to the Markov NumberTree 
The fundamental solutions are
1    X:  16    Y:  8    Z:  8 
1    X:  18    Y:  12    Z:  6 
1    X:  25    Y:  20    Z:  5 
1    X:  9    Y:  9    Z:  9 
2    X:  8    Y:  4    Z:  4 
2    X:  9    Y:  6    Z:  3 
3    X:  3    Y:  3    Z:  3 
3    X:  6    Y:  4    Z:  2 
4    X:  4    Y:  2    Z:  2 
5    X:  5    Y:  4    Z:  1 
6    X:  3    Y:  2    Z:  1 
8    X:  2    Y:  1    Z:  1 
9    X:  1    Y:  1    Z:  1 
jagy@phobeusjunior:~$

Notice how, for each $n > 1,$ we can create a solution with $n=1,$ from $(n;x,y,z)$ to $(1,nx,ny,nz).$ This may help proving something, as we may demand $(1;a,b,c)$ and hope to prove the list full. 
