Q:

A conjecture involving the equation $x^n+y^n+z^n= (x+y)^n$

Given $x,y,z,n\in\mathbb{N}$,  $x,y,z,n>0$, and $x\neq y$, my conjecture is that

$$
(x+y+z)^n-(x+z)^n-(y+z)^n=0 \Longrightarrow x^n+y^n+z^n-(x+y)^n=0,
$$

where one can easily recognize Fermat's equation among the hypotheses. 
My ultimate goal is to find the structural connection between these two equations, without supplementary knowledge (e.g. we know that $(x+y+z)^n-(x+z)^n-(y+z)^n=0\Rightarrow n\leq 2$). This means, for instance, to find a way to transform the first equation into the second one, e.g. by an appropriate change of variables, showing that the two equations admit the same solutions. Hence, the conjecture.
However, I tried to prove such statement by means of the triangular inequality and the binomial expansion, also by reductio ad absurdum. But I could not find a way through this, therefore

Can you suggest some idea or technique to prove or disprove such statement, without using the fact that we know that $(x+y+z)^n-(x+z)^n-(y+z)^n=0\Rightarrow n\leq 2$?

This post is clearly linked to this one Literature about the equation $x^m+y^m+z^m=(x+y)^m$.
NOTE: This post is a correction of a previous one, that I deleted because of errors pointed out by a kind user. Sorry for the inconvenience!
Thank you!

A:

Write the implication as $L=0 \Rightarrow R=0$. Then $L+R=\sum_{1 \leq i \leq n-1}\binom{n}{i}z^{n-i}[(x+y)^i-x^i-y^i]$, which is greater than zero for $n>2$ because the term $(x+y)^i-x^i-y^i$ is positive for $i >1$. Thus, $L=0 \Rightarrow R>0$ for $n>2$, without any additional knowledge.

