nt.number theory - Prime divisors of numbers 2^n + 3

(Edited as the comments below suggest)

The ABC conjecture seemed to me like it would play a roll, however it comes up a little short:

"Are there infinitely many primes $p$ so that for each $p$ there is some integer $n$ with $p^2|2^n + 3?"$

If the ABC conjecture is true, then this answer to this question is almost "no", but still there is a problem at the end of the argument.

The ABC conjecture states that for any $epsilon > 0$ there is a constant $K_epsilon$ so that for any co-prime triple $A < B < C$ with $A+B = C$ then
$$C le K_epsilonprod_{p|ABC}p^{1 + epsilon}.$$

So, if there is such an infinite collection of primes, then for the corresponding infinite $n$ where this is true then $2^n + 3 = p^2C$ then
$$p^2C le K_epsilon(6Cp)^{1+epsilon}.$$

(Edited: The following sentence is incorrect "But this will clearly run into problems for sufficiently large $p.$" But I wanted to leave it so Kevin's comment makes sense.)

Note that as $C = C(p)$ is a function of $p$ then the $C^epsilon$ (when $C$ is square-free, or nearly square-free) term may still allow this inequality to work.