Proof: Let $p=8k-1, k \in \mathbf{N}_{+}, p$ be a prime number, and the symbol
$\left(\frac{-1}{p}\right)=1 \Leftrightarrow$ odd prime $p \equiv 1(\bmod 4)$;
$\left(\frac{2}{p}\right)=1 \Leftrightarrow$ odd prime $p \equiv \pm 1(\bmod 8)$.
From the given, we know $p=7(\bmod 8)$, hence
$$\left(\frac{-1}{p}\right)=-1 \text{, and }\left(\frac{2}{p}\right)=1 \text{. }$$

But $p!\left(2^{n}+1\right)$, i.e., $2^{n} \equiv-1(\bmod p)$.
Since $\left(\frac{x}{p}\right)$ is a completely multiplicative function, we have
$$-1=\left(\frac{-1}{p}\right)=\left(\frac{2^{n}}{p}\right)=\left(\frac{2}{p}\right)^{n}=1^{n}=1 \text{. }$$

Contradiction.