Any given prime $P \in \mathbb P$ is either of the form $$ P \equiv -1 \mod 6$$ or $$ P \equiv 1 \mod 6.$$ In other words, every prime greater than 3 can be expressed in the form $$P = 5+6n \vee 7 + 6n $$ Proof is easy. Any prime must be
$$\begin{cases} P \equiv 1 \mod 2 \\ P \equiv 1 \vee -1 \mod 3 \end{cases}$$
if the latter is $1$, we get $2 \ | \ (P-1) \wedge 3 \ | \ (P-1),$ so $6 \ | \ (P-1)$ and if it is $-1$, we get $2 \ | \ (P+1) \wedge 3 \ | \ (P+1),$ so $6 \ | \ (P+1)$.
Now, we also know that there are infinitely many primes of both forms, as proven somewhere on this site.
This means that there is infinite number of such numbers $n \in \mathbb N$ that $$5+6 n \in P$$ and infite number of such $k \in \mathbb N$ that $$7+6 k \in P.$$ This also means that we have $\forall P>3 \in \mathbb P,$ some kind of infinite set $S_P$ such that $$P + 6 s \in \mathbb P, \forall s\in S_P $$ because, say, we assume $P = 5+ 6 m$ then $$(5 + 6m) + 6s = 5+(m+s)6$$ and here we use the first argument, just shifted forwards by $m$.
Now, denote the set of such $k$ for some $P$ that $P + 6k \in \mathbb P$ by $K_P$.
If we keep all $P \equiv 5 \mod 6$, I would suggest that the intersection $K_{P_1}\cap K_{P_2}$ of two such sets is also infinitely large. Is it?
So if we $P_1 = 5 + 6s$ and $P_2 = 5 + 6 t$, we get two sets $\{s+1,s+2,s+3\cdots\}$ and $\{m+1,m+2,m+3\cdots\}.$ Is it possible that the running terms never coincide with respect to primality of the numbers in question? Could this be proven impossible with, for example, using pigeonhole principle? Or something to do with filters? I am sure that if the definitions were modified a bit, one could acquire filter bases.