, 2011

Let $k$ be an arbitrary integer. Note that
$$(9a^2k^3)^2 + 3 = 3(27a^4k^6 + 1)$$
and
$$(9a^3k^4)^3 - a = a(3^6a^8k^{12} - 1) = a(27a^4k^6 - 1)(27a^4k^6 + 1)$$
It follows that for every $k \in \mathbb{Z}$ the number $27a^4k^6 + 1$ is a common divisor of the numbers $n^2 + 3$ and $m^3 - a$ with $n = 9a^2k^3$ and $m = 9a^3k^4$. So it is enough to prove that there are infinitely many primes $p$ such that $p \mid 27a^4k^6 + 1$ for some integer $k$.
Suppose that there are only finitely many such primes and these are $p_1, p_2, \dots, p_r$. If we take $k = p_1p_2\dots p_r + 1$, then it is clear that the number $27a^4k^6 + 1$ is not divisible by any $p_i$ for $1 \le i \le r$ and that it is also greater than 1. It follows that it has a prime divisor $p$, which is different from every $p_i$ for $1 \le i \le r$. We have obtained a contradiction, which finishes the proof. $\square$