Let $a$, $b$ and $k$ be positive integers. I want to prove that the only solution of the system $$\left\{\begin{array}{rclr} 2k^2+1&=&a^2&(1)\\ 6k^2+1&=&b^2&(2) \end{array} \right.$$ is $(k,a,b)=(2,3,5)$.
I checked by computer that there was no solution with $3\leq k\leq10^8$.
I know that the values of $k$ which satisfy $(1)$ are the terms of the sequence $(k_n)$ defined by $$\left\{\begin{array}{l} k_0=2\\ k_1=12\\ k_{n+2}=6k_{n+1}-k_n \end{array} \right.$$ and that the values of $k$ which satisfy (2) are those of the sequence $(k'_n)$ defined by $$\left\{\begin{array}{l} k'_0=2\\ k'_1=20\\ k'_{n+2}=10k'_{n+1}-k'_n \end{array} \right.$$
but I cannot prove that if $m\geq1$ and $n\geq1$, then $k_n\neq k'_m$.
Let $a$, $b$ and $k$ be positive integers such that \begin{eqnarray} 2k^2+1&=&a^2,\\ 6k^2+1&=&b^2. \end{eqnarray} Then clearly $k$ is coprime to $a$ and $b$ and $$a^2+(2k)^2=b^2,$$ so $(a,2k,b)$ is a primitive Pythagorean triple. This means there exist coprime integers $m>n>0$, not both odd, such that \begin{eqnarray} a&=&m^2-n^2,\\ k&=&mn,\\ b&=&m^2+n^2. \end{eqnarray} Plugging this back into the first equation of the original system we see that $$1=m^4-4m^2n^2+n^4,$$ which is a Thue equation, for which effective methods exist. PARI/GP tells me that the unique integral solution with $m>n>0$ is $(m,n)=(2,1)$, corresponding to $a=3$, $k=2$, $b=5$.