Given positive integer $k$, define the subset $S(k)$ of positive integers $n$ where for every odd prime power $p^k < n$, $n - p^k$ is prime. In other words, $$S(k) = \{n \in \mathbb{N} \mid \forall p \in \mathbb{P} : (p > 2 \wedge p^k < n) \Rightarrow n - p^k \in \mathbb{P}\}.$$
Question: Is $S(k)$ is finite for every $k > 0$?
I suspect this is the case from initial computer searches. I am hoping there is an elementary proof of this (if it's true), but I have had little success.
Some observations: any $n < 3^k$ will be in $S(k)$ since the condition holds vacuously true on $n$. Furthermore, the largest odd element of $S(k)$ must be $\leq 3^k + 2$ as otherwise it could be written as $3^k + \text{composite even number}$ and thus not be in $S(k)$. So, it suffices to focus attention on the even elements of $S(k)$.
Intuitively, it makes sense that the larger you go, the less likely a number belongs to $S(k)$ since there are more "chances" for at least one $n - p^k$ to be composite. But it is not at all obvious that this is actually "impossible to belong to $S(k)$ once we exceed a certain number".