Can we have $x^n\equiv (x+1)^n \pmod m$ for large enough $n$? $x^n\equiv (x+1)^n$ For what values of m and n can we find an x that solves this?

It is not clear from your question what is fixed and what varies. To illustrate: suppose you want a solution with $n=10$ --- is that a large enough $n$? Very well, then --- pick your favorite $x$, say, $x=42$. Then calculate $43^{10}-42^{10}$ and call it $Q$. Then if $m$ is $Q$, or any factor of $Q$, you will have $(x+1)^n\equiv x^n\pmod m$. 
If that's not what you want, please edit your question to clarify. 
EDITing in things from the comments: 


*

*For fixed $n\gt3$ it is probably hard to find a useful characterization of those $m$ for which there is a solution. 

*For fixed $m$, let $n=\phi(m)$, then $x^n\equiv(x+1)^n\equiv1\pmod m$ provided $x$ and $x+1$ are relatively prime to $m$. 

*There are (lots of) solutions not of the form of item 2. 
