For any integer $n$, let $n! = 2^kd$ for some non-negative integer $k$ and odd integer $d$.
For any power of $2$ , i.e. $n=2^h$, where $h$ is non-negative integer. $$ \begin{aligned} k & =\left[\frac{n}{2}\right]+\left[\frac{n}{2^2}\right]+\cdots+\left[\frac{n}{2^{h-1}}\right]+\left[\frac{n}{2^h}\right]\\ & =2^{h-1}+2^{h-2}+\cdots+1 \\ & =2^h-1 \\ & =n-1 \end{aligned} $$ $\therefore 2^{n-1} | n !$ for any $n=2^h$.
We now deal with odd integers $n>1$. Noting that $[\frac{n}{2^r}] =[\frac{n-1}{2^r}]$ we have $$\begin{aligned}k & =\left[\frac{n}{2}\right]+\left[\frac{n}{2^2}\right]+\left[\frac{n}{2^3}\right]+\cdots \\& =\left[\frac{n-1}{2}\right]+\left[\frac{n-1}{2^2}\right]+\cdots\\ &<(n-1)\left(\frac{1}{2}+\frac{1}{2^2}+\cdots\right)\\&=n-1 \cdots (*)\end{aligned} \tag*{} $$
Therefore $k<n-1$ and hence $2^{n-1}\not|n!$ for any odd integer $n$.
Now for any even integer $n$ which is not a power of $2$, we let $n=2^he$ for some positive integer $h\geq 1$ and odd integer $e\geq 3$. Then
$$\displaystyle \begin{aligned}k & =\left[\frac{2^h e}{2}\right]+\left[\frac{2^h e}{2^2}\right]+\cdots+\left[\frac{2^h e}{2^h}\right]+\left[\frac{e}{2}\right]+\left[\frac{e}{2^2}\right]+\cdots \\\\ & < (2^{h-1} e+2^{h-2} e+\cdots+e)+(e-1) \quad \textrm{ (Using (*))} \\& =\frac{e\left(2^h-1\right)}{2-1}+e-1\\&=2^h e-1\\&=n-1\end{aligned}\tag*{} $$
Therefore $k<n-1$ again and hence $2^{n-1}\not|n!$ for any even integer $n$ which is not a power of $2$.
We may now conclude that $n=2^h$ are the positive integers satisfying $2^{n-1}|n!,$ where $h$ are non-negative integer.
Do we have a simpler way to find all the positive integers satisfying $2^{n-1}|n!$?