Consider the set $p_0=f(1)$, $p_k$=$f(kp_0+1)$. Using the fact that $a-b|f(a)-f(b)$ in $\mathbb{Z}[x]$, $kp_0|p_k-p_0$ so $p_0|p_k$. If $f(n)$ is prime for all n, then $p_i$ is prime for all $i$, so $p_0=p_k$ for all k, but this implies that the polynomial $f(x)-p_0$ has an infinity of roots, contradicting the fact that it is a polynomial (which must have a finite degree).

Here is a solution that constructs a slightly different contradiction. Suppose FTSOC, $f(n)$ is prime for all $n\in\mathbb{Z}^{+}$. We know that $(a-b)|(f(a)-f(b))$ for $a,b\in\mathbb{Z}$. WLOG, let $b=1$ and let $f(b)=p$, $p$ a prime. Then we have $(a-1)|(f(a)-p)$. Now, take $a\in\{jp+1\}_{j=1}^{j=k}$. Then, we have $jp|(jq_{j}-p)$ for all $1\le j\le k$ for arbitrary primes $q_1, \ldots, q_k$. Since $q_1$ is prime and $p|(q_1-p)$, it follows that $q_1=\pm p$. Similarly, $q_j = \pm p$ for all $1\le j\le k$. Thus, by pigeonhole we get $f(x)=f(y)$ for every pair of integers $(x,y)$, which means $f$ is constant, contradiction.

$f(n)|f(n+kf(n))$