Suppose $3x+1=k^2$ and $6x-2=m^2$, where $k, m\in\mathbb{N}_{0}$ Then $2(9x^2-1)=k^2m^2\Longleftrightarrow 3(6x^2-1)=k^2m^2-1=(km-1)(km+1)$, Obviously $6x ^2-1\geq 5>3$. Given that $6x^2-1$ is prime, then $km-1\in\{1, 3\}$. iff $km-1=3\Longleftrightarrow km=4$ and $km+1=6x^2-1\Longrightarrow 6x^2=6\Longleftrightarrow x= 1$, iff $km-1=1\Longleftrightarrow km+1=3=3(6x^2-1)\Longleftrightarrow x^2=\dfrac{1}{3}$. So $x=1$

$3x+1 = k^2$,$(6x-2)=t^2$ $(k^2t^2-1) = 3p$ $k^2t^2 \equiv 1 \mod 3$ implicating $kt=3n+1,3n+2 , n \geq 0$ $(3n)(3n+2) = 3p$ $n(3n+2)=p$ $n=1,3n+2=5$ $(3n+1)(3n+3) = 3p$ $(3n+1)(n+1)=p$ $n+1=1,3n+1=1$ contradiction $kt=4$ which give $k=2,t=2,x=1$