We will prove that such numbers do not exist.

The quadratic equation $x^2 - qx + 1 = 0$ has two rational roots $t$ and $\frac{1}{t}$ for which $t + \frac{1}{t} = q$. It easily follows by induction that $a_m = t^m + \frac{1}{t^m}$. Let's assume that there exist numbers $b_0, b_1, \dots, b_n$ satisfying the condition of the problem.

Lemma. Let $f(x) = c_n x^n + c_{n-1} x^{n-1} + \dots + c_1 x + c_0$ be a polynomial with nonzero integer coefficients for which $c_{n-k} = c_k$ for each $k = 0, 1, \dots, n$ and $\sum_{i=0}^n c_i = 0$. Then $f(x) = (x-1)^2 g(x)$, where $g(x)$ is a polynomial with integer coefficients.

Proof: From the condition we have that $f(1) = 0$. As
$$f'(x) = n c_n x^{n-1} + (n-1) c_{n-1} x^{n-2} + \dots + c_1,$$
it follows that
$$2 f'(1) = (n c_n + (n-1) c_{n-1} + \dots + c_1) + (n c_0 + (n-1) c_1 + \dots + c_{n-1}) = n(c_n + c_{n-1} + \dots + c_1 + c_0) = 0$$
and therefore $x = 1$ is a double root. The lemma is proved.

The polynomial $f(x) = b_n x^{2n} + b_{n-1} x^{2n-1} + \dots + b_1 x^{n+1} + 2b_0 x^n + b_1 x^{n-1} + b_2 x^{n-2} + \dots + b_{n-1} x + b_n$ satisfies the conditions of the lemma. It's not hard to see that
$$t^n \left( b_n \left( t^n + \frac{1}{t^n} \right) + b_{n-1} \left( t^{n-1} + \frac{1}{t^{n-1}} \right) + \dots + 2b_0 \right) = f(t) = (t-1)^2 g(t).$$
Moreover, if $t = \frac{r}{s}$, $(r, s) = 1$, then from $\frac{r}{s} + \frac{s}{r} = q > 3$ it easily follows that $r \ge s + 2$, i.e. $r - s \ge 2$. Then $g\left(\frac{r}{s}\right)$ is a number of the form $\frac{l}{s^{2n-2}}$.

Finally $b_0 a_0 + b_1 a_1 + \dots + b_n a_n$ is presented in the form $\frac{(r-s)^2 l}{r^n s^n}$ and because the number $r-s \ge 2$ and $(r-s, r) = (r-s, s) = 1$, then the numerator will always square a prime number. $\square$