First we shall prove that $a_n^2$ is the difference of two consecutive cubes. To do so, we shall prove by induction that $a_n^2 = (b_n + 1)^3 - b_n^3$.

When $n = 1$, this is true.

Suppose for $n \ge 1$, this is true. We have
$$\begin{aligned} (b_{n+1} + 1)^3 - b_{n+1}^3 &= 3b_{n+1}^2 + 3b_{n+1} + 1 \\ &= 3(4a_n + 7b_n + 3)^2 + 3(4a_n + 7b_n + 3) + 1 \\ &= 48a_n^2 + 147b_n^2 + 168a_n b_n + 84a_n + 147b_n + 37 \\ &= (7a_n + 12b_n + 6)^2 + (3b_n^2 + 3b_n + 1) - a_n^2 \\ &= a_{n+1}^2 + (b_n + 1)^3 - b_n^3 - a_n^2 = a_{n+1}^2 \end{aligned}$$
where the last equality is by induction hypothesis.