Let $p$ be an odd prime where $p \mid n$. Then,
$$\begin{aligned} 4^n + 1 &= (4^{\frac{n}{p}})^p + 1 = (4^{\frac{n}{p}} + 1)((4^{\frac{n}{p}})^{p-1} - (4^{\frac{n}{p}})^{p-2} + \dots - 4^{\frac{n}{p}} + 1) \\ &= (4^{\frac{n}{p}} + 1)x \end{aligned}$$

where $x = (4^{\frac{n}{p}})^{p-1} - (4^{\frac{n}{p}})^{p-2} + \dots - 4^{\frac{n}{p}} + 1 > 1$. So, we have
$$\begin{aligned} m^{4n+1} - 1 &= m^{(4^{\frac{n}{p}}+1)x} - 1 \\ &= (m^{4^{\frac{n}{p}}+1} - 1)(m^{(4^{\frac{n}{p}}+1)(x-1)} + m^{(4^{\frac{n}{p}}+1)(x-2)} + \dots + m^{4^{\frac{n}{p}}+1} + 1). \end{aligned}$$
Since $m^{4n+1} - 1$ is a prime and $x > 1$, $m^{4^{\frac{n}{p}}+1} - 1 = 1$. Thus, $m^{4^{\frac{n}{p}}+1} = 2$, and so $m = 2$ and $4^{\frac{n}{p}}+1 = 1$. Therefore, $\frac{n}{p} + 1 = 0$, which is a contradiction. We conclude that $n$ has no odd prime factors, hence is a power of 2.