I'm interested in the following problem: do there exist infinitely many prime numbers $p$ such that $p^2|2^{n}+3$ for some natural number $n$?

Some motivation:

If we replace the function $2^n + 3$ with the $f(n)$ where $f \in \mathbb{Z}[x]$ is non-constant that this is true (follows Hensel lemma). So, it's rather natural to try proving this for some other non-polynomial functions. $2^n + 3$ is an easy example of such function. There is also another good reason: sequence $a_n = 2^{n} + 3$ satisfies the reccurence relation: $a_{n+2} = 3a_{n+1} - 2a_{n}$. And for example this problem is true for Fibonacci sequence. So, for Fibonacci it's easier even if the closed form of Fibonacci numbers is more complicated. But I think that the reason of this is that the Fibonacci numbers satisfy some "good" identities which other sequences don't have to share.

Now some remarks:

It's easy exercise to prove that there are infinitely many primes $p$ such that $p|2^{n}+3$. Also, if we try "correcting" $n$ to work also for $p^2$ and we try $m=n+k(p-1)$ we see that it is possible unless $p$ is Wieferich prime, i.e. satisfies $p^2|2^{p-1}-1$. And this gives us nothing as we don't know much about Wieferich primes...

This method can of course be generalized in such way: if $p|2^{n}+3$ and order of $2$ mod $p^2$ is greater than order of $2$ mod $p$ then we can find $m$ such that $p^2|2^{m}+3$. But I don't really think that it helps.

I'm interested in some information about this problem (especially if it's open or not) and also related problems. We can ask a general question: for which functions $f$ we know that this is true?

Edit: Sorry for the confusion with $k$, deleted.