Let $a_n=\displaystyle\sum_{k=0}^{n}2^{3k}\binom{2n+1}{2k+1}$ and $b_n=\displaystyle\sum_{k=0}^{n}2^{3k}\binom{2n+1}{2k}$. Then, $$\begin{align*} \sqrt{8}a_n + b_n =\ &\sum_{k=0}^{n}\binom{2n+1}{2k+1}(\sqrt{8})^{2k+1}+ \sum_{k=0}^{n}\binom{2n+1}{2k} (\sqrt{8})^{2k} \\ \\=\ &\sum_{k=0}^{2n+1}\binom{n}{k} (\sqrt{8})^{k}\\ \\=\ & (\sqrt{8}+1)^{2n+1} \end{align*}$$

Similarly, we have $$\sqrt{8}a_n - b_n =(\sqrt{8}-1)^{2n+1}$$

Multiplying these two equations gives $$8a_n^2 - b_n^2 = 7^{2n+1} \implies 8a_n^2 = b_n^2 + 7^{2n+1}$$

Now, the claim is equivalent to showing that $(b_n^2 + 7^{2n+1})$ is not a multiple of $5$. Firstly, $$7^{2n+1}=49^n\times 7\equiv (-1)^n\times 2 \equiv \pm 2\pmod{5}$$

In order for $(b_n^2 + 7^{2n+1})$ to be a multiple of $5$, $b_n\equiv \pm 2\pmod{5}$ must hold. However, a square number cannot end with $2$ or $3$. Hence, it is impossible.