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?

Besides Vieta jumping one can successfully use Pell's equation. For $n=1$ this is demonstrated here, giving infinite families of solutions. This may work for other $n$, too. The author, Titu Andreescu, has written several notes on such Diophantine equations. If you search his articles, in particular on quadratic Diophantine equations, there are more references.
Remark: For Vieta jumping, for the different equation $x^2+y^2+z^2=nxyz$, see this question on MSE.
