Does $(x+y)^m=x^m+y^m+z^m$ imply $(x+y+z)^m=(x+z)^m+(y+z)^m$? Let $x,y,z,m\in\mathbb{N}$, and $x,y,z,m>0$, and also $x>y$. 
My problem is to understand if, under these sole hypotheses, we can prove that

$(x+y)^m=x^m+y^m+z^m \Longrightarrow (x+y+z)^m=(x+z)^m+(y+z)^m.$

If yes, how can we prove it? 
If not, which other hypotheses are needed, in order to make the implication true?
EDIT: I am also interested in the softer versions of the statement, i.e. whether we can prove or not that

$(x+y)^m=x^m+y^m+z^m \Longrightarrow (x+y+z)^m\lessgtr (x+z)^m+(y+z)^m,$

and, if not, which additional conditions we need to make the statement(s) true.

Expanding the given condition using binomial we get,
$$ z^m = {m\choose{1}} xy^{m-1} + {m\choose{2}} x^2y^{m-2} +...........+{m\choose{m-1}} x^{m-1}y $$
Now if we expand the claim,  we get,
$$
\sum \frac{m!}{a!b!c!} x^ay^bz^c = z^m + \Bigg( {m\choose{1}} xz^{m-1} + {m\choose{2}} x^2z^{m-2} +...........+{m\choose{m-1}} x^{m-1}z \Bigg)  + \Bigg( {m\choose{1}} yz^{m-1} + {m\choose{2}} y^2z^{m-2} +...........+{m\choose{m-1}} y^{m-1}z \Bigg)
$$ 
where $a, b, c$ are natural numbers such that $a+b+c = m$ and $a,b,c \neq m$.
We already know the value of $z^m$ so we can substitute that, 
$$
\sum \frac{m!}{a!b!c!} x^ay^bz^c = \Bigg(  {m\choose{1}} xy^{m-1} + {m\choose{2}} x^2y^{m-2} +...........+{m\choose{m-1}} x^{m-1}y  \Bigg) + \Bigg( {m\choose{1}} xz^{m-1} + {m\choose{2}} x^2z^{m-2} +...........+{m\choose{m-1}} x^{m-1}z \Bigg)  + \Bigg( {m\choose{1}} yz^{m-1} + {m\choose{2}} y^2z^{m-2} +...........+{m\choose{m-1}} y^{m-1}z \Bigg)
$$ 
We can simplify this to, 
$$
\sum \frac{m!}{a!b!c!} x^ay^bz^c = \Bigg(  {m\choose{1}} ( xy^{m-1} + yz^{m-1} + zx^{m-1} ) +
 {m\choose{2}} (x^2y^{m-2} + y^2z^{m-2} + z^2x^{m-2})...........+{m\choose{m-1}}( x^{m-1}y^ + y^{m-1}z + z^{m-1}x ) \Bigg) 
$$ 
I can't see how to prove that $LHS=RHS$ from here.  In fact, since you are not sure whether your claim was true, this seems to indicate that the claim is false.
Can anyone try something from here?
Edit As the other answer says the claim is clearly contradicting FMT.
