4. Find all pairs of positive integers $(k, n)$, satisfying
$$
k!=\left(2^{n}-1\right)\left(2^{n}-2\right)\left(2^{n}-4\right) \cdots\left(2^{n}-2^{n-1}\right) .
$$
4. Let the right-hand side of the equation be $R_{n}$. Then
$$
v_{2}\left(R_{n}\right)=\sum_{i=0}^{n-1} v_{2}\left(2^{n}-2^{i}\right)=\sum_{i=0}^{n-1} i=\frac{1}{2}(n-1) n \text {. }
$$

It is well-known that $v_{2}(k!)=k-S_{2}(k)$, where $S_{2}(k)$ represents the sum of the digits of $k$ in its binary representation. Hence,
$$
\begin{array}{c}
k-S_{2}(k)=\frac{1}{2}(n-1) n \\
\Rightarrow k=\frac{1}{2}(n-1) n+S_{2}(k) \\
\geqslant \frac{1}{2}(n-1) n+1 .
\end{array}
$$

By directly checking $n=1,2,3,4$, we obtain two solutions:
$$
(k, n)=(1,1),(3,2) \text {. }
$$

For $n \geqslant 5$, we will prove:
$$
\left(\frac{1}{2}(n-1) n+1\right)!>2^{n^{2}} \text {. }
$$

Since $R_{n} \leqslant\left(2^{n}\right)^{n}=2^{n^{2}}$, it follows that $k!>R_{n}$ for $n \geqslant 5$, and there are no positive integer pairs $(k, n)$ that satisfy the condition.
For $n=5$, we directly verify equation (1):
$$
11!=39916800>2^{25}=33554432 \text {. }
$$

For $n>5$,
$$
\begin{array}{l}
\left(\frac{1}{2}(n-1) n+1\right)!=11!\times \prod_{i=12}^{\frac{1}{2}(n-1) n+1} i \\
>2^{25} \times 8^{\frac{1}{2}(n-1) n-10} \\
=2^{25+\frac{3}{2}(n-1) n-30} \\
\geqslant 2^{\frac{3}{2}(n-1) n-5} \geqslant 2^{n^{2}} .
\end{array}
$$

In summary, all the positive integer pairs that satisfy the condition are
$$
(k, n)=(1,1),(3,2) .
$$