We will prove that if $x_{n+1}$ is a perfect square of a rational, then $x_n$ is also a perfect square of a rational, and the desired result is obtained by a simple induction.

Note first that since $3$ doesn't divide $b$, it will not divide any of the denominators of the sequence terms.

From the recursive relation we have that $x_m = 3x_{m-1}^3 + x_{m-1}$. Setting $x_{m-1} = \frac{p}{q}$ where $q$ is not divisible by $3$ (*) and $(p,q)=1$, then
$$x_m = 3x_{m-1}^3 + x_{m-1} = \frac{3p^3 + pq^2}{q^3} = \frac{p(3p^2 + q^2)}{q^3}.$$
Since $(p,q)=1$, this is the reduced form of $x_m$. Indeed, the numbers $p(3p^2+q^2)$, $q^3$ are coprime, since if $s$ is a prime dividing both of them, then $s|q^3 \Rightarrow s|q$ and $s|3p^2+q^2$ so $s|3p^2$. But $s$ doesn't divide $p$, so $s|3 \Rightarrow s=3$, $3|q$, absurd due to (*).

Moreover $x_m$ is a perfect square so both numerator and denominator in the reduced form should be perfect squares.
Since the denominator is a perfect square, $q$ is a perfect square, let it be $q = a^2$. For the numerator, we have $p(3p^2+q^2) = \kappa^2$, so both of them should be perfect squares since they are coprime.

Therefore, $p = b^2$, $3p^2 + q^2 = c^2$ and $x_{m-1} = \frac{p}{q} = \frac{b^2}{a^2}$, so $x_{m-1}$ is a perfect square of a rational.

Similarly, going back $x_{m-2}$ is also a perfect square of a rational, and so on, till we arrive at $x_1$.