There are infinitely many primes that divide $2^{n^3+1}-3^{n^2+1}+5^{n+1}$? $$q(n)=2^{n^3+1}-3^{n^2+1}+5^{n+1}$$
Claim: If $p>2$ divides $q(n)$ then $p$ not divides $q(n+k)$, for $k>0$
If $p|q(n)$, then $$p|q(n)(5^k)(5^{k^2+2nk+k^2})(2^{k^3+3n^2k+3nk^2})=2^{(n+k)^3+1}(a)-3^{(n+k)^2+1}(b)+5^{n+k}(c)$$
But we supposed that it's true for $a=b=c=1$, instead $a,b,c>1$
Since a prime that divides $q(n)$ does not divide any $q(n + k)$, we conclude that for every $q(n)$ there is always a different prime. So there are infinitely many primes that divide $q(n)$
Do you know if the proof is correct? I am not very sure
 A: The equation is, for integers $n$,
$$q(n)=2^{n^3+1}-3^{n^2+1}+5^{n+1} \tag{1}\label{eq1A}$$
As stated in aschepler's comment comment, your claim that any prime $p \gt 2$ which divides \eqref{eq1A} does so for only one value of $n$ is incorrect.
Instead, let $S$ be the set of odd primes $p$ which divide $q(n)$ for at least one integer $n$. Note that $S$ is non-empty, e.g., $q(1) = 4(5)$ so $5 \in S$, and $q(2) = 2(197)$ so $197 \in S$. However, as lulu's comment indicates, $3 \not\in S$ since odd $n$ results in $q(n) \equiv 1 + 1 \equiv 2 \pmod{3}$, and even $n$ gives $q(n) \equiv 2 + 5 \equiv 1 \pmod{3}$.
Assume $S$ is finite, i.e., it has some $m \gt 1$ elements of $p_i$ for $1 \le i \le m$. Let
$$n_1 = \prod_{i=1}^{m}(p_i - 1) \tag{2}\label{eq2A}$$
This means that
$$n_1^3 + 1 \equiv n_1^2 + 1 \equiv n_1 + 1 \equiv 1 \pmod{p_i - 1} \; \forall \; 1 \le i \le m \tag{3}\label{eq3A}$$
Using Fermat's little theorem, and that $5^{n_1+1} \equiv 5 \pmod{5}$, we have
$$q(n_1) \equiv 2 - 3 + 5 \equiv 4 \pmod{p_i}  \; \forall \; 1 \le i \le m \tag{4}\label{eq4A}$$
Next, since $n_1$ is even,
$$q(n_1) \equiv -3 + 5 \equiv 2 \pmod{8} \tag{5}\label{eq5A}$$
i.e., $q(n_1)$ has exactly one factor of $2$. However, since $q(n_1) \gt 2$, it must then have at least one odd prime factor, but \eqref{eq4A} shows this is not in $S$, contradicting that $S$ contains all such odd prime factors. This contradiction means the assumption of $S$ being finite is false, i.e., there are infinitely many primes which divide $q(n)$ in \eqref{eq1A} for at least one value of $n$.
A: As a side note, you can do much better, providing a lower bound for the number of prime factors $f(n)$ dividing $2^{n^3+1}-3^{n^2+1}+5^{n+1}$ for infinitely many $n$.
Indeed, by Theorem 1 here, there exist infinitely many $n\in \mathbf{N}$ and a real $\theta>1$ such that
$$
f(n) \ge \mathrm{slog}_\theta(n),
$$
where $\mathrm{slog}_\theta$ is the "superlogarithm" in base $\theta$, i.e., the largest integer $k$ such that $\theta^{\theta^{\theta^{\dots}}}\le n$, where $\theta$ is repeated $k$ times.
A: For $p$ prime and $n \in \mathbb{Z}$, let $M(p,n)$ be the largest integer so that $p^{M(p,n)}$ divides $n$.
Suppose that only finitely many primes divide $q$ namely $z_{1},....,z_{m}$. Set $Z= \{z_{1},...,z_{m}\}$. Note that some subset $H \subseteq Z$ is ``maximal'' in the sense that $\exists n_{H} \in \mathbb{N}$ so that $\left(\prod_{w \in H} w\right) | q(n_{H})$ and we have
$$H \subsetneq \mathbb{} G \subseteq Z \text{ }\Rightarrow  \forall z \in \mathbb{Z}\text{ } \prod_{w \in G}w \not \div q(z) $$
Thus for all $\beta \in \mathbb{Z}$, $q(n_{H} + \beta \phi(n_{H}))$ is a product of primes in $H$. Moreover for $\beta$ sufficiently large we have $M(h, q(n_{H} + \beta \phi(n_{H})) = M(h, q(n_{H}))$ whenever $h \in H$ (here we take into consideration if $h$ were to be $2,3$ and $5$ and apply Euler's theorem). Thus we have $q(n_{H} + \beta\phi(n_{H})) \leq q(n_{H})$ for all $\beta$ sufficiently large which is impossible.
