Four, (50 points) Find all positive integers $n$, such that $2^{n}+2n$ is a factorial of a positive integer.

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Let $2^{n}+2 n=m!$.
First, consider the case where $m \geqslant 8$.
Let $v_{2}(x)$ denote the exponent of the prime factor 2 in the prime factorization of the positive integer $x$.
Let $\alpha=v_{2}(n)$.
Clearly, $n>3$.
Since $v_{2}(m!)>3$, then
$$
\begin{array}{l}
v_{2}(2 n)>3 \Rightarrow \alpha+1>3 \Rightarrow \alpha>2 \\
\Rightarrow n \geqslant 2^{\alpha}=(1+1)^{\alpha}>1+\alpha \\
\Rightarrow v_{2}\left(2^{n}+2 n\right)=1+\alpha \\
\Rightarrow v_{2}(m!)=1+\alpha .
\end{array}
$$

Let $[x]$ denote the greatest integer less than or equal to the real number $x$.
$$
\begin{array}{l}
\text { Also, } v_{2}(m!)=\left[\frac{m}{2}\right]+\left[\frac{m}{4}\right]+\left[\frac{m}{8}\right]+\cdots \\
\geqslant\left(\frac{m}{2}-\frac{1}{2}\right)+\left(\frac{m}{4}-\frac{3}{4}\right)+1=\frac{3 m-1}{4},
\end{array}
$$

Thus, $\alpha \geqslant \frac{3 m-5}{4}, n \geqslant 2^{\frac{3 m-5}{4}}$.
We will now prove that $2^{\frac{3 m-5}{4}} \geqslant(m-5)(m-1)$ for $m \geqslant 8$.
Indeed, when $m=8$,
$$
\begin{array}{l}
2^{\frac{3 m-5}{4}}=2^{4 \frac{3}{4}}>16 \sqrt{2} \geqslant 21 \\
=(m-5)(m-1) .
\end{array}
$$

Assume that $2^{\frac{3 m-5}{4}} \geqslant(m-5)(m-1)$, we need to prove:
$$
\begin{array}{l}
2^{\frac{3(m+1)-5}{4}} \geqslant(m-4) m \\
\Leftarrow 2^{\frac{3}{4}} \geqslant \frac{(m-4) m}{(m-5)(m-1)} .
\end{array}
$$

Indeed, for $m \geqslant 8$,
$$
\begin{array}{l}
\frac{(m-4) m}{(m-5)(m-1)} \\
=\left(1+\frac{1}{m-5}\right)\left(1+\frac{1}{m-1}\right) \\
\leqslant\left(1+\frac{1}{3}\right)\left(1+\frac{1}{7}\right)=\frac{32}{21} \\
2^{n} \geqslant 2^{2^{\frac{3 m-5}{4}}} \geqslant 2^{(m-5)(m-1)} \text {. } \\
\text { Since } 2^{m-5}=(1+1)^{m-5} \\
\geqslant \mathrm{C}_{m-5}^{0}+\mathrm{C}_{m-5}^{1}+\mathrm{C}_{m-5}^{m-6}+\mathrm{C}_{m-5}^{m-5} \\
=1+(m-5)+(m-5)+1 \\
=2 m-8 \geqslant m \\
\Rightarrow 2^{n}+2 n>\left(2^{m-5}\right)^{m-1} \geqslant m^{m-1}>m!.
\end{array}
$$

Therefore, when $m \geqslant 8$, there are no positive integers $n$ that satisfy the condition.

For $m<8$, it is easy to verify that only $m=4$ has $n=4$ that satisfies the condition.

In conclusion, only $n=4$ makes $2^{n}+2 n$ a factorial of a positive integer.