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

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.
