If we substitute $a_n = 2x_n - 1$ in the $2x_n - 1 = 2(2x_{n-1}x_{n-2} - x_{n-1} - x_{n-2} + 1) - 1 = (2x_{n-1} - 1)(2x_{n-2} - 1)$ we get $a_n = a_{n-1}a_{n-2}$. Initial conditions of the recurrence relation are $a_1 = 2 \cdot 2 - 1 = 3$, $a_0 = 2a - 1$. It is required that $a_0 = 2a - 1$ must be a perfect square. Since $a_0 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2a - 1$, we get $a = 2k^2 + 2k + 1$.

Now let us prove that if $\forall k \in \mathbb{Z}: a = 2k^2 + 2k + 1$ then $\forall n \in \mathbb{N}$, $2x_{3n} - 1$ are perfect squares.

Moreover it is possible to prove for $\forall n \in \mathbb{N}$ terms $a_{3n}$ are perfect squares and terms $a_{3n+1}$, $a_{3n+2}$ are of the form $3s^2$, $s \in \mathbb{Z}$.

Let's proceed by induction. Base of induction is trivial.

If $a_{3n} = k^2$ and $a_{3n+1} = 3m^2$ then $a_{3n+2} = k^2 \cdot 3m^2 = 3(km)^2$.

If $a_{3n+1} = 3k^2$ and $a_{3n+2} = 3m^2$ then $a_{3n+3} = (3km)^2$.

If $a_{3n+2} = 3k^2$ and $a_{3n+3} = m^2$ then $a_{3n+4} = 3(km)^2$.

Thus we conclude for the numbers $a = 2k^2 + 2k + 1$, $\forall k \in \mathbb{Z}$ all terms of the progression with the form $a_{3n} = 2x_{3n} - 1$ are perfect squares.