If $|f(0)|>0$, f is a monotonic periodic function and thus is constant; however, the functional equation admits no constant solutions so we must have
$f(0)=0\;\;\;(1)$.
Now, substituting $x=0$ into the functional equation gives us
$f(f(y))=y^n\;\;\;(2).$
Note that (2) allows to write the functional equation as
$f(x+f(y))=f(x)+y^n=f(x)+f(f(y))\;\;\;(3).$
Also note that by the functional equation and (1), the range of $f$ contains all non-negative numbers (fix $x=0$ and vary y). Hence any non-negative number may be substituted for f(y).
Thus the functional equation can be written as
$f(x+y)=f(x)+f(y),x\in\mathbb{R}, y\geq0\;\;\;(4)$
Since f is monotonic, we must have that for $x\geq0, f(x)=Cx$ for some constant C.
Substituting this partial solution into the original functional equation, we see that this is only possible if $n=1$; that is, if $n\neq1$ the functional equation does not admit solutions.
If $n=1$, f is clearly surjective by the original functional equation and thus (4) holds $\forall x,y\in\mathbb{R}.$
It follows that the only potential solutions are of the form $f(x)=Cx$ for some constant C (monotonic additive functions are linear).
Substituting into the functional equation, we find that the only solutions are $f(x)=\pm x\;\;\;(\forall x\in\mathbb{R})$.