We show the stronger result that the set of prime $p \equiv \pm 5 \pmod{12}$ such that $p \mid 9^n - 2$ for some $n$ (since $9 = 3^2$, we can work with $9$ instead of $3$, by restricting to even exponents), but $p \nmid 2^m - 3$ for any $m$, is infinite.

Note that if $p \mid 9^n - 2$ for some $n$, then $9^{nm} \equiv 2^m \equiv 3 \pmod p$ would imply that $3$ is a quadratic residue mod $p$, which contradicts $p \equiv \pm 5 \pmod{12}$. (The latter can be easily checked using the law of quadratic reciprocity.)

Thus, it suffices to justify that the set of primes $p \equiv \pm 5 \pmod{12}$ that divide $9^n - 2$ for some $n$ is infinite. Suppose these primes are finite, say $p_1, \dots, p_k$ (there is at least one such prime, for instance, $p = 7$ for $n = 1$). Then, for $n = \prod_{i=1}^k (p_i - 1) > 1$, we have $9^n \equiv 1 \pmod{p_i}$ by Fermat's little theorem, which implies $p_i \nmid 9^n - 2$ for each $i$. However, $9^n - 2 \equiv 7 \pmod{12}$ must have a prime divisor $q \equiv \pm 5 \pmod{12}$ (if all prime factors were $\pm 1 \pmod{12}$, their product with multiplicities yields $9^n - 2 \equiv \pm 1 \pmod{12}$, a contradiction), distinct from each $p_i$. The desired result follows. $\square$