Does this sequence always give a square number?

Question : Supposing that a sequence $\{a_n\}$ is defined as $${a_{n+3}}^2=-{a_{n+2}}^2+2{a_{n+1}}^2+48a_{n+1}a_{n}+32{a_n}^2\ (n\ge 1)$$ $$a_1=a_2=a_3=1$$ then, is $a_n$ a square number for any $n$?

For example, we can see $$\sqrt{a_n} : 1,1,1,3,1,5,7,3,17,11,23,45,1,91,89,93,271,85,457,627,287,1541,967,2115,\cdots$$

Motivation : I found the following question in a book without any proof.

Supposing that a sequence $\{b_n\}$ is defined as $$b_{n+3}=-b_{n+2}+2b_{n+1}+8b_n\ (n\ge 1)$$ $$b_1=b_2=b_3=1,$$ then, prove that $b_n$ is a square number for any $n$.

This is obvious by the following relational expression : $$(b_{n+3}-b_{n+2})^2=64b_{n+1}b_n,$$ which can be shown by induction on $n$.

After solving this question, I've tried to find a similar sequence by using computer. Then, I reached the above expectation. The expectation seems true, but I can neither find any counterexample nor prove that the sequence always gives a square number. Can anyone help?

Let $b_n = -b_{n-1} - 2b_{n-2}$, $b_1=1, b_2=-1$.

I claim that $a_n = b_n^2$, which is obviously a perfect square, for all $n \in \Bbb{Z}^+$. (See footnote below)

Proof by induction:

Base case: $a_1 = a_2 = a_3 = b_1^2 = b_2^2 = b_3^2 = 1$

Hypothesis: $a_n = b_n^2$ for $n \le k$

Induction step: Start by using the induction hypothesis on the expression for $a_{k+1}$:

$$\begin{align} &a_{k+1}^2 = -a_k^2+2a_{k-1}^2+48a_{k-1}a_{k-2}+32a_{k-2}^2 = -b_k^4 + 2b_{k-1}^4+48b_{k-1}^2b_{k-2}^2+32b_{k-2}^4 \end{align}$$

We want to show that $a_{k+1}^2 = b_{k+1}^4$. Expanding $b_{k+1}^4$ gives:

$$\begin{align} &b_{k+1}^4 = (-b_k - 2b_{k-1})^4 = b_k^4 + 8b_k^3b_{k-1}+24b_k^2b_{k-1}^2+32b_kb_{k-1}^3+16b_{k-1}^4 = \\\\ &-b_k^4+2b_{k-1}^4+48b_{k-1}^2b_{k-2}^2+32b_{k-2}^4 + \\ &{\color{red}{2b_k^4+8b_k^3b_{k-1}+24b_k^2b_{k-1}^2+32b_kb_{k-1}^3+14b_{k-1}^4-48b_{k-1}^2b_{k-2}^2-32b_{k-2}^4}} \end{align}$$

If we now can show that the red part equals zero, we're done. This can be done by replacing $b_k$ by $(-b_{k-1}-2b_{k-2})$ and expanding.

Motivation for my claim: A search on OEIS.

EDIT: I just realized that proof by induction is completely unnecessary. Showing that the expansion of $b_{n}^4$ is equal to the recursion formula for $a_n^2$ is enough, i.e. the details in the induction step above. I guess I was too involved in the algebraic manipulations to see the big picture at the time.