Let $L(z)$ be the smallest prime divisor of $z$.

Lemma. If $a$, $b$, $c \in \mathbb{N}$ such that $a \neq b$ we have
$$L\left(\frac{a^c - b^c}{a - b}\right) \geq L(abc)$$
Proof. Assume the contrary, then there should be an integer $p$ such that $\gcd(p, a) = \gcd(p, b) = 1$ and $p \mid \frac{a^c-b^c}{a-b}$. Now consider two cases, if $a \equiv b \pmod{p}$ then by lifting the exponent lemma $\nu_p(a^c-b^c) = \nu_p(a-b)$ which is a contradiction. If $a \not\equiv b \pmod{p}$ then $ord_p(ab^{-1}) \neq 1$ ($b^{-1}$ is multiplicative inverse of $b$ modulo $p$). But we know that $ord_p(ab^{-1}) \mid c$ and we have $p > \text{ord}_p(ab^{-1}) > L(c)$ which is a contradiction.

Consider a set
$$A_k = \{x \in \mathbb{N} \mid L(x) \geq k\}$$
now if $a$, $b$, $c \in A_k$ and $d \mid \frac{a^c-b^c}{a-b}$ then by lemma we have $L(d) < L(abc) < n$ and hence $d \in A_k$. Now it is enough to take $k > n$. ■