Here is a proof for $n \geq 1$, unfortunately for $n<0$ we get a weaker ienquality.
For all $n \notin(0,1)$ we have
$(y+\frac{x}{n})^n = y^n(1+\frac{x}{ny})^n > y^n(1+\frac{x}{y})$
with the last inequality following from Generalized Bernolli
For $n \geq 1$ you also have:
$ y^n+xy^{n-1}> y^n+x^n =z^n\,,$
and by combining these two you get the desired result.
Maybe someone can fix it in general.
Added: Inspired by only's answer.
If $n=-1$, your equation is
$\frac{1}{x}+\frac{1}{y}=\frac{1}{z} \Rightarrow xz+yz+=xy \Rightarrow z= \frac{xy}{x+y}$
You need $y-x >z$, or equivalently
$y-x > \frac{xy}{y+x} \Leftrightarrow y^2 > x^2+xy$
Note that $y=3, x=2$ is a counterexample to this equation, thus $y=3, x=2, z=\frac{6}{5}, n=-1$ is a counterexample to your claim.
In general, if $n=-m<0 $ you want to prove that
$\frac{1}{x^m}+\frac{1}{y^m} =\frac{1}{z^m} \mbox{implies} \, y-\frac{x}{m} >z$
but since
$\frac{z^m}{y^m}+\frac{z^m}{x^m}=1\,,$ if you set $\frac{z^m}{y^m} =\frac{1}{2}+\alpha$ and $\frac{z^m}{x^m} =\frac{1}{2}-\alpha$, the desired inequality becomes
$\frac{1}{\sqrt[m]{\frac{1}{2}+\alpha}}-\frac{1}{m\sqrt[m]{\frac{1}{2}-\alpha}} >1$ Since this is increasing in $\alpha$, is easy to see that this is true for all $0 < \alpha < \frac{1}{2}$ if and only if
$\frac{m-1}{m\sqrt[m]{\frac{1}{2}}} =\lim_{\alpha \to 0^+} \frac{1}{\sqrt[m]{\frac{1}{2}+\alpha}}-\frac{1}{m\sqrt[m]{\frac{1}{2}-\alpha}} \geq 1$
or equivalently
$2 \geq (\frac{m}{m-1})^m $.
But this is not possible, since the RHS is a sequence which is decreasing to $e$. This shows that for each $m$ there exists some $\alpha$ which makes the inequality fail....