It is possible to prove elementarily that there are infinitely many primes that divide some element of the sequence $a\_0 = k\ge 0$, $a\_n = a\_{n-1}^2+ 1$ for all $n\ge 1$ by showing that for all $m$, there exists $C$ that depends only on $m$ s.t. $(a\_n, a\_{m + n})\le C$ and then showing that this is not possible if there are only finitely many primes that divide it. 


However, I don't know of any results that might help prove the following, stronger statement:


For any (finite) set of primes $S$, there are only finitely many solutions to 
$$n^2 + 1 = m$$
for $m,n$ s.t. $p|mn \Rightarrow p\in S$


How would one prove this?



Prime factors of the members of a certain recurrence
It is even true that the equation $n^2+1=m$ with $p|m\Rightarrow p\in S$ has only finitely many solutions. To see this note that every $m$ satisfying this property can be written as $m=m\_1m\_2^3$, where $m\_1$ has only prime factors in $S$ and is cubefree. In particular $m\_1$ comes from a finite set of integers. Hence it suffices to show that for a given $m\_1$ the number of integral solutions of $n^2+1=m\_1m\_2^3$ is finite. But this is a special case of a result by Siegel stating that an elliptic curve has only finitely many integral points.


