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?
 A: 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.
A: 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. 
A: We only care about finding $n$, as opposed to finding all (families of) solutions. 
This solution follows that of a similar problem $(x+y+u+v)^2 = n^2 xyuv$ that was posed in Vietnam Mathematical Olympiad 2002. 

The given equation is equivalent to
$$ x^2 +  (2y + 2z - n yz) x  + (y+z)^2 = 0 $$
Let $n$ be a integer with positive integer solutions.
Let $(x_0, y_0, z_0) $ be a solution with the minimum sum $x+y+z$.
WLOG $x_0 \geq y_0 \geq z_0$.   
Observe that
1. $x_0 \mid ( y_0 + z_0)^2$.
2. $x_0$ is a positive integer solution to the quadratic function $f(x)  x^2 + (2 y_0 + 2z_0 - n y_0z_0 ) x + (y+z)^2 $.
3. The other solution to the equation $f(x) = 0$ is $x_1 =  -(2 y_0 + 2z_0 - n^2 y_0z_0 ) - x_0 = \frac{ (y_0+z_0)^2 } { x_0 }$.
4. So $ (x_1, y_0, z_0)$ is a solution, and thus $ x_1 \geq x_0$.
5. By considering the shape of the parabola, since $y_0 \not \in (x_0, x_1)$, hence $f(y_0) \geq  0 $.   
Expanding, we get:   
$0 \leq f(y_0) = y_0^2 + (2y_0 + 2z_0 - n y_0z_0)y_0 + (y_0 + z_0)^2$
$ \leq y_0^2 + (2y_0 + 2y_0)y_0 + (y_0 + y_0)^2 - n y_0^2 z_0$
$ = 9 y_0^2 - ny_0^2 z_0$ 
So $ n z_0 \leq 9$, which implies $ n \leq 9$.   
Now, that we have a bound on $n$, it remains to find which of them have possible solutions. 
For now, this is left as an exercise to the reader (Sorry!)

Thinking out loud. How can we show that $n = 7 $ has no solutions?
No results yet.
Suppose we had a solution to $ (x+y+z) ^2 = 7 xyz$.
Then $ x + y + z = 7 k$, so $ 7k^ 2 = xyz$.
If  7 divides two of these numbers, then it must divide all three.
Let $ x = 7x', y = 7y', z = 7 z'$,and we have $ ( x' + y' + z')^2 = 49 x'y'z'$, but we know from above that no solutions exist.
Hence, WLOG, $ 7 \mid x,  7 \not \mid yz, 7 \mid y+z$.
Let $ x = 7 x ', y + z = 7a$.
We have $ (x'+a)^2 = x'y(7a-y) $.     
