Let $P(x,y)$ be the assertion $f(x)+f(y) \mid (x+y)^{k}$.
(1) $f(1)=1$
Proof:First assume that $f(1) \neq 1$.$P(1,1)$ implies $f(1)=2^{a}$ for some non-negative integer $a$.
$P(2,2)$ implies $f(2)=2^{b}$ for some non-negative integer $b$.
$P(1,2)$ implies that $f(1)+f(2) \mid 3^{k}$.If both $f(a),f(b)$ are even,then we get a contradiction as an even integer doesn't divide an odd integer.So,one of them is odd i.e. equals to $1$.As we assumed that $f(1) \neq 1$,so,$f(2)=1$.
So,$2^{a}+1=3^{r}$........(i),then by Zsigmondy's Theorem,$a=1$ or $3$.
Now,$P(1,3)$ implies that $f(1)+f(3) \mid 4^{k}$.So,$f(1)+f(3)=2^{i}$ where $i$ is a positive integer.Since $f(1)$ is a power of $2$,so,$f(1)=f(3)$.
Now,$P(2,3)$ implies that $1+2^{a} \mid 5^{k}$ $\Rightarrow$ $2^{a}+1=5^{r}$.......(ii).From (i),we know that $a=1$ or $3$,but none is valid for (ii),a contradiction.So our first assumption was wrong and $f(1)=1$.
(2) $f(p-1)=p-1$
Proof:$P(1,p-1)$ implies that $f(p-1)=p^{s}-1$.
$P(p-1,p-1)$ implies that $f(p-1)=p^{s}-1 \mid 2^{k-1}(p-1)^{k}$.So ,the set of distinct prime divisors of $p-1$ and $p^{s}-1$ is same.But from Zsigmondy's Theorem,we know that this can be possible if and only if $s=1$ i.e. $f(p-1)=p-1$.
(3) $f(n)=n$ for all $n \in \mathbb{N}$
Proof:$P(n,p-1)$ implies that $f(n)+p-1 \mid (n+p-1)^{k}$.
Now,$f(n) \equiv -(p-1)$ $\text{(mod}$ $f(n)+p-1)$.
$\Rightarrow$ $n-f(n) \equiv n+p-1$ $\text{(mod}$ $f(n)+p-1)$.
$\Rightarrow$ $(n-f(n))^{k} \equiv (n+p-1)^{k} \equiv 0$ $\text{(mod}$ $f(n)+p-1)$
Here the value of $(n-f(n))^{k}$ is fixed for each particular $n$.But if $p \to \infty $, $(f(n)+p-1) \to \infty$.Therefore,$(n-f(n))^{k}=0$ and $f(n)=n$ for all $n \in \mathbb{N}$.