Solution:

From the definition of the sequence we see that $\left(a_{n+1} - 2a_n\right)^2 = 3a_n^2 + 1$. After simplification we get
$$a_{n+1}^2 + a_n^2 - 4a_n a_{n+1} = 1$$
Adding $3a_{n+1}^2$ to both sides of the last equation gives us $\left(2a_{n+1} - a_n\right)^2 = 3a_{n+1}^2 + 1$ and after taking the square root of both sides we get
$$\sqrt{3a_{n+1}^2 + 1} = 2a_{n+1} - a_n$$
On the other hand, from the definition of the sequence we see that
$$\sqrt{3a_{n+1}^2 + 1} = a_{n+2} - 2a_{n+1}.$$
From (1) and (2) we conclude that $a_{n+2} = 4a_{n+1} - a_n$ which together with $a_1 = 1$, $a_2 = 4$ implies that all terms are integers.