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?

At the risk of presenting a trivial example: we can even have it for small enough $m$. Just let $m=1$. Any integer (mod $1$) is equivalent to zero.
$x^n \equiv (x+1)^n$ (mod $1$).
EDIT: Your conjecture also holds if $n=0$. $x^0 = (x+1)^0$.
