This will follow with very small $k$ from Schinzel's hypothesis H.

Let $p$ be of the form $12n+1$. There are no congruence obstructions $p$ and $p_1=(p-1)/12=n$ and $p_2=(p+1)/2=6n+1$ to be simultaneously prime. By the above conjecture, they are prime infinitely often answering with very small $k$ since $p^2-1=12\cdot2\cdot p_1 p_2$

This will follow with very small $k$ from Schinzel's hypothesis H or as @Gerry Myerson points out Dickson's conjecture.

Let $p$ be of the form $12n+1$. There are no congruence obstructions $p$ and $p_1=(p-1)/12=n$ and $p_2=(p+1)/2=6n+1$ to be simultaneously prime. By the above conjecture, they are prime infinitely often answering with very small $k$ since $p^2-1=12\cdot2\cdot p_1 p_2$

This will follow with very small $k$ from the k tuple conjecture and the stronger Schinzel's hypothesis H.

Let $p$ be prime of the form $12n+1$. There are no congruence obstructions $(p-1)/12=n$ and $(p+1)/2=6n+1$ to be simultaneously prime. By the above conjectures, they are prime infinitely often answering with very small $k$.

This will follow with very small $k$ from Schinzel's hypothesis H.

Let $p$ be of the form $12n+1$. There are no congruence obstructions $p$ and $p_1=(p-1)/12=n$ and $p_2=(p+1)/2=6n+1$ to be simultaneously prime. By the above conjecture, they are prime infinitely often answering with very small $k$ since $p^2-1=12\cdot2\cdot p_1 p_2$