6. Let $q=2^{m+1}+1$, by the condition we know $3^{2^{m}} \equiv-1(\bmod q)$, hence $3^{2^{m+1}} \equiv 1(\bmod$ $q)$. This indicates $\delta_{q}(3) \mid 2^{m+1}$, but $\delta_{q}(3) \times 2^{m}$. Therefore, $\delta_{q}(3)=2^{m+1}$.

On the other hand, by Euler's theorem, we know $3^{\varphi(q)} \equiv 1(\bmod q)$, so $2^{m+1} \mid \varphi(q)$, i.e., $(q-1) \mid \varphi(q)$. Combining $\varphi(q) \leqslant q-1$, we know $\varphi(q)=q-1$, thus $q$ is a prime.