Question: For $k>3$ does there exist an odd prime $q_k$ such that $p_k=2^kq_k+1$ is prime and $p_k$ divides $a_k=\dfrac{3^{2^{k-1}}+1}{2}$?\

If $k=3$ the answer is Yes because for $q_3=5$ we get $p_3=a_3=41$. \

$a_4=3281=17\cdot 193$ but neither $17=2^4\cdot 1+1$ nor $193=2^4\cdot 12+1$ qualifies to be $p_4$ because $1$ and $12$ are not (odd) prime numbers.\

$a_5,a_6$ and $a_7$ turn out to be prime numbers, so, the answer to the question is No (see the recursive definition below). $a_8$ has $61$ digits and none of its factors qualifies to be $p_8$. Unfortunately $a_9$ has approximately $121$ digits and $a_{10}$ has approximately 243 digits. I would like to see a proof for the No answer for $k>3$ or see the condition on $k$ for which the answer is Yes.\

There is also a recursive definition for $a_k$ : $a_2=5,a_k=2^k(a_2\cdots a_{k-1})+1,k>2,$ which makes it very clear why $k=3$ is a Yes answer. ($a_1$ doesn't really matter but a meaningful definition for it is $a_1=2.$)