hint

The brutal approach works. The characteristic polynomial for the sequence is $ \lambda^2 - 14\lambda + 1 = 0$, with roots $ 7 \pm 4\sqrt {3} = (2 \pm \sqrt {3})^2$. Therefore, after determining the coefficients (from the first two terms), the general formula is $ a_n = \frac {1} {4} \left ( (2 + \sqrt {3})^{2n - 1} + (2 - \sqrt {3})^{2n - 1} \right )$. It follows $ 2a_n - 1 = \frac {1} {2} \left ( \sqrt {(2 + \sqrt {3})^{2n - 1}} - \sqrt {(2 - \sqrt {3})^{2n - 1}} \right )^2$, since $ (2 + \sqrt {3})(2 - \sqrt {3}) = 1$. But $ (2 + \sqrt {3})^{2n - 1} = \frac {1} {2^{2n - 1}} (\sqrt {3} + 1)^{2(2n - 1)} = 2 \left ( \frac {1} {2^{n}} (\sqrt {3} + 1)^{2n - 1} \right )^2$ and $ (2 - \sqrt {3})^{2n - 1} = \frac {1} {2^{2n - 1}} (\sqrt {3} - 1)^{2(2n - 1)} = 2 \left ( \frac {1} {2^{n}} (\sqrt {3} - 1)^{2n - 1} \right )^2$. Therefore $ 2a_n - 1 = \left ( \frac {(\sqrt {3} + 1)^{2n - 1} - (\sqrt {3} - 1)^{2n - 1}} {2^n} \right )^2$. It remains to check the quantity squared is an integer. Denote $ A = \sqrt {3} + 1$ and $ B = \sqrt {3} - 1$. We have $ A^2 + B^2 = 8$ and $ A^2B^2 = 4$. But $ A^{2n + 1} - B^{2n + 1} = (A^2 + B^2)(A^{2n - 1} - B^{2n - 1}) - A^2B^2(A^{2n - 3} - B^{2n - 3})$, hence $ \frac {A^{2n + 1} - B^{2n + 1}} {2^{n + 1}} = 4\frac {A^{2n - 1} - B^{2n - 1}} {2^n} - \frac {A^{2n - 3} - B^{2n - 3}} {2^{n - 1}}$, so it is proven by induction. In fact we have proven that $ 2a_n - 1 = b_n^2$, with the sequence $ (b_n)_{n\geq 0}$ given by $ b_0 = - 1$, $ b_1 = 1$ and $ b_{n + 1} = 4b_n - b_{n - 1}$. (This particular relation is prone to many developments and has been often used; see also TST-Romania 2006).

Thanks for your solution mavropnevma.

Shoki, I would like to see your solution using Pell Equation.

put $ 2a_n=b_n$ then consider this pell equation : $ x^2-3y^2=1$ as u can see we have $ b_1^2-3(1)^2=1$ and also $ b_2^2-3(15)^2=1$ and also ... since $ (x_0,y_0)=(2,1)$ is the fundamental answer of this pell equation,and also the formula which gives all other solutions is $ x_{n+1}=2x_n+3y_n \ , \ y_{n+1}=x_n+2y_n$ we have $ 2 | x_{2k}$ after finding the general formula for the sequence $ b_n$ and also the general formula for the sequence $ x_{2k}$ we can easily understand that $ b_n=x_{2k}$ (maybe u can prove this also by induction!) so for all $ n$ we have $ b_n^2-3y_{2k}^2=1$ which gives $ (b_n-1)(b_n+1)=3y_{2k}^2$ since $ b_n$ is even so we get $ \gcd(b_n-1,b_n+1)=1$ and so there exists $ (r,s) \in N^2$ such that $ b_n-1=r^2 \ , \ b_n+1=3s^2 \ , rs=y$ and so we get what we wanted. (note that the case $ b_n-1=3r^2 \ , \ b_n+1=s^2 \ , rs=y$ is easily omitted since $ 3r^2=s^2-2$ which is a contradiction(modulo $ 3$))

The first few values of $2a_n-1$ for $n\ge 2$ are $5^2,19^2,71^2,265^2$. Consider the sequence $b_1=1,b_2=5$ and $b_n=4b_{n-1}-b_{n-2}$ for $n\ge 3$. Indeed the first few values are $5,19,71,265$. We will prove by induction that $2a_n-1=b_n^2$, clearly the first few values work. Not let us suppose we have $2a_i-1=b_i^2$ for all $i\le n$. This means $a_n=\frac{b_n^2+1}{2},a_{n-1}=\frac{b_{n-1}^2+1}{2}$. Now $a_{n+1}=14a_n-a_{n-1}=14\left(\frac{b_n^2+1}{2}\right)-\frac{1}{2}-\frac{b_{n-1}^2}{2}=7b_n^2-\frac{b_{n-1}^2}{2}+\frac{13}{2}$. So we aim to prove $\frac{b_{n+1}^2+1}{2}=7b_n^2-\frac{b_{n-1}^2}{2}+\frac{13}{2}$. Now \begin{align*}\frac{b_{n+1}^2+1}{2} & =7b_n^2-\frac{b_{n-1}^2}{2}+\frac{13}{2}\\ \iff b_{n+1}^2+1 & =14b_n^2-b_{n-1}^2+13\\ \iff (4b_n-b_{n-1})^2 & = 14b_n^2-b_{n-1}^2+12\\ \iff b_n^2+b_{n-1}^2& =4b_nb_{n-1}+6\end{align*} after simplifying. To prove $b_n^2+b_{n-1}^2 =4b_nb_{n-1}+6$ we will again use induction. Indeed it is true for the first few values of $n$. Now assuming $b_n^2+b_{n-1}^2 =4b_nb_{n-1}+6$ we see: \begin{align*}b_n^2+b_{n-1}^2& =4b_nb_{n-1}+6\\ 16b_n^2-8b_nb_{n-1}+b_{n-1}^2+b_n^2 & =16b_n^2-4b_nb_{n-1}+6\\ (4b_n-b_{n-1})^2+b_n^2 & =4(4b_{n}-b_{n-1})b_n+6\\ b_{n+1}^2+b_n^2 & =4b_{n+1}b_n+6\end{align*} This completes both inductions, so that $2a_n-1=b_n^2$ is a perfect square for all $n\ge 2$. And clearly $2a_0-1=2a_1-1=1$ are squares, so it is true for $n\ge 0$ also.