Let $(a_n)$ be the A001921 sequence

$$ a_0 = 0,\ a_1 = 7, \quad a_{n+2} = 14a_{n+1} - a_n + 6. $$

Let $(b_k)$ be the (almost)"tower-of-squares" sequence defined by

$$ b_0=2, \quad b_{k+1}=2b_k^2-1 $$

Is it true that $a_{2^kn+2^{k-1}-1}$ is always divisible by $b_k$, for any $k,n\geq 0$ ?

I have checked this up to $k=6$. For example :

$a_{2n}$ is always divisible by $b_0=2$.

$a_{4n+1}$ is always divisible by $b_1=7$.

$a_{8n+3}$ is always divisible by $b_2=97$.

Etc. Up to : $a_{128n+65}$ is always divisible by $b_6=2011930833870518011412817828051050497$.

This is a cross-post from a MSE question.