Cesenatico

Solution:

Suppose that the set $S$ of primes that divide integers of the form $a_{n}=2^{n^{3}+1}-3^{n^{2}+1}+5^{n+1}$ is finite. We will find a contradiction by exhibiting an integer $n$ such that $a_{n}$ possesses at least one prime factor that does not belong to $S$.

Let $n$ be the product of all the numbers $p-1$ as $p$ ranges over $S$ (that is, $n$ is Euler's totient function of the product of the primes in $S$). Since $a_{1}=2^{2}-3^{2}+5^{2}=20=2^{2} \cdot 5$, then $4=5-1$ divides $n$. Moreover, for every $p \in S$, except possibly the cases $p=2,3,5$, we have, by Fermat's little theorem, $2^{n} \equiv 3^{n} \equiv 5^{n} \equiv 1(\bmod p)$, from which
$$a_{n}=2^{n^{3}+1}-3^{n^{2}+1}+5^{n+1} \equiv 2-3+5 \equiv 4 \quad(\bmod p) \quad \forall p \in S, p \neq 2,3,5$$
None of the primes in $S$ can therefore divide $a_{n}$, except possibly 2, 3 and 5. On the other hand $a_{n}$ is not divisible by 3: indeed, since $n$ is even, we have $a_{n} \equiv 2-0+2 \equiv 1(\bmod 3)$. Moreover $a_{n}$ is not divisible by 5 either: indeed, since $n$ is a multiple of $4$, $a_{n} \equiv 2-3+0 \equiv -1(\bmod 5)$. If, for contradiction, $a_{n}$ had only prime factors belonging to $S$, then $a_{n}$ would necessarily have to be a power of 2. Since $n \geq 2$, we have $0<3^{n^{2}+1}-5^{n+1}<2^{n^{3}}$, from which
$$2^{n^{3}}=2^{n^{3}+1}-2^{n^{3}}<a_{n}<2^{n^{3}+1}.$$
It follows that $a_{n}$ lies between two consecutive powers of 2, and therefore cannot itself be a power of 2; this contradiction gives the thesis.