In the articleAnalytic solutions and integrability for bilinear recurrences of order six Andrew Hone considered general Somos-6 recurrence $$\tau_{n+6}\tau_{n}=\alpha\tau_{n+5}\tau_{n+1}+\beta\tau_{n+5}\tau_{n+2}+\gamma\tau_{n+3}^2$$ with arbitrary coefficients $\alpha$, $\beta$, $\gamma$. He gave explicit analytic solution in the form $$\tau_n=AB^n\dfrac{\sigma(\mathbf{v}_0+n\mathbf{v})}{\sigma(\mathbf{v})^{n^2}},$$ where $\mathbf{v}$, $\mathbf{v}_0\in\mathbb{C}^2$ and $\sigma$ is a Kleinian sigma function associated with some genus $2$ curve $\mu^2=4\nu^5+c_3\nu^3+c_2\nu^2+c_1\nu+c_0.$ From Baker's addition formula (see Baker's An introduction to the theory of multiply periodic functions (1907)) $$\dfrac{\sigma(\mathbf{u}+\mathbf{v}) \sigma(\mathbf{u}-\mathbf{v})}{\sigma(\mathbf{u})^2\sigma(\mathbf{v})^2}= \wp_{22}(\mathbf{u})\wp_{12}(\mathbf{v})-\wp_{12}(\mathbf{u})\wp_{22}(\mathbf{v})+ \wp_{11}(\mathbf{v})-\wp_{11}(\mathbf{u})$$ follows that for some fumctions $f_k$, $g_k$ ($1\le k\le 4$) $$\tau_{m+n}\tau_{m-n}=\sum\limits_{k=1}^{4}f_k(m)g_k(n).$$ It means that infinite matrix consisting from $A_{mn}=\tau_{m+n}\tau_{m-n}$ has rank at most $4$ and every minor of order $5$ vanishes.
Let's apply this theory to the given sequence. This minor corresponds to two $5$-tuples of $m$'s and $n$'s $(m,5,4,3,2)$ and $(n,4,2,1,0)$ $$\Delta=\left| \begin{array}{ccccc} \tau_{m-n} \tau_{m+n} & \tau_{m-4} \tau_{m+4} & \tau_{m-2} \tau_{m+2} & \tau_{m-1} \tau_{m+1} & \tau_m^2 \\ \tau_{5-n} \tau_{n+5} & 97 & 11 & 10 & 4 \\ \tau_{4-n} \tau_{n+4} & 25 & 5 & 2 & 4 \\ \tau_{3-n} \tau_{n+3} & 22 & 2 & 2 & 1 \\ \tau_{2-n} \tau_{n+2} & 10 & 2 & 1 & 1 \\ \end{array} \right|=0.$$ We can take $m=n$ if we want to find $\tau_{2n}$ and $m=n+1$ for $\tau_{2n+1}$. But it is necessary to divide by $$\left| \begin{array}{cccc} 97 & 11 & 10 & 4 \\ 25 & 5 & 2 & 4 \\ 22 & 2 & 2 & 1 \\ 10 & 2 & 1 & 1 \\ \end{array} \right|=2\cdot 9.$$ We know from David Speyer's answer that $9$ is not important for us. The only problem is $2$. ($\tau_0=\tau_1=1$ are not a problem too.) But $$\left| \begin{array}{ccccc} \tau_{m-n} \tau_{m+n} & \tau_{m-4} \tau_{m+4} & \tau_{m-2} \tau_{m+2} & \tau_{m-1} \tau_{m+1} & \tau_m^2 \\ \tau_{5-n} \tau_{n+5} & 97 & 11 & 10 & 4 \\ \tau_{4-n} \tau_{n+4} & 25 & 5 & 2 & 4 \\ \tau_{3-n} \tau_{n+3} & 22 & 2 & 2 & 1 \\ \tau_{2-n} \tau_{n+2} & 10 & 2 & 1 & 1 \\ \end{array} \right|\equiv (\tau_{m+2}\tau_{m-2}+\tau_{m+4}\tau_{m-4})(\tau_{4+n}\tau_{4-n}+\tau_{5+n}\tau_{5-n})\pmod{2}$$ must be $0\pmod{2}$. So we have the expression $$0=\Delta=18\tau_{m+n}\tau_{m-n}+(\tau_{m+2}\tau_{m-2}+\tau_{m+4}\tau_{m-4})(\tau_{4+n}\tau_{4-n}+\tau_{5+n}\tau_{5-n})+2\cdot (\mathrm{Some Polynomial})$$ which can be divided by $2\tau_{m-n}=2$ (because $m=n$ or $m=n+1$) in order to get $\tau_{m+n}$.