We have $(2+\sqrt{3})^{4}=97+56 \sqrt{3}=14(7+4 \sqrt{3})-1=14(2+\sqrt{3})^{2}-1$. Hence $(2+\sqrt{3})^{k+2}=14 (2+\sqrt{3})^{k}-(2+\sqrt{3})^{k-2}$. Thus if $(2+\sqrt{3})^{k}=a_{k}+b_{k} \sqrt{3}$, then $a_{k+2}=14 a_{k}-a_{k-2}$.

Now suppose the sequence $c_{k}$ satisfies $c_{1}=1, c_{2}=5, c_{k+1}=4 c_{k}-c_{k-1}$. We claim that $c_{k}^{2}-c_{k-1} c_{k+1}=6$. Induction on $k$. We have $c_{3}=19$, so $c_{2}^{2}-c_{1} c_{3}=25-19=6$. Thus the result is true for $k=2$. Suppose it is true for $k$. Then $c_{k+1}=4 c_{k}-c_{k-1}$, so $c_{k+1}^{2}=4 c_{k} c_{k+1}-c_{k-1} c_{k+1}=4 c_{k} c_{k+1} -c_{k}^{2}+6=c_{k}(4 c_{k+1}-c_{k})+6=c_{k} c_{k+2}+6$, so the result is true for $k+1$.

Now put $d_{k}=c_{k}^{2}+1$. We show that $d_{k+2}=14 d_{k+1}-d_{k}$. Induction on $k$. We have $d_{1}=2, d_{2}= 26, d_{3}=362=14 d_{2}-d_{1}$, so the result is true for $k=1$. Suppose it is true for $k$. We have $c_{k+3}- 4 c_{k+2}+c_{k+1}=0$. Hence $12+2 c_{k+3} c_{k+1}-8 c_{k+2} c_{k+1}+2 c_{k+1}^{2}=12$. Hence $2 c_{k+2}^{2}-8 c_{k+2} c_{k+1}+2 c_{k+1}^{2}=12$. Hence $16 c_{k+2}^{2}-8 c_{k+2} c_{k+1}+c_{k+1}^{2}+1=14 c_{k+2}^{2}+14-c_{k+1}^{2}-1$, or $(4 c_{k+2}-c_{k+1})^{2}+ 1=14(c_{k+2}^{2}+1)-(c_{k+1}^{2}+1)$, or $c_{k+3}^{2}+1=14(c_{k+2}^{2}+1)-(c_{k+1}^{2}+1)$, or $d_{k+3}=14 d_{k+2}-d_{k+1}$. So the result is true for all $k$.

But $a_{1}=2, a_{3}=26$ and $a_{2k+3}=14 a_{2k+1}-a_{2k-1}$, and $d_{1}=2, d_{2}=26$ and $d_{k+1}=14 d_{k}-d_{k-1}$. Hence $a_{2k-1}=d_{k}=c_{k}^{2}+1$.