, 2019

We will get an upper bound on $n$ from the speed at which $v_{2}\left(L_{n}\right)$ grows.
From
$$L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right) \cdots\left(2^{n}-2^{n-1}\right)=2^{1+2+\cdots+(n-1)}\left(2^{n}-1\right)\left(2^{n-1}-1\right) \cdots\left(2^{1}-1\right)$$
we read
$$v_{2}\left(L_{n}\right)=1+2+\cdots+(n-1)=\frac{n(n-1)}{2} .$$
On the other hand, $v_{2}(m!)$ is expressed by the Legendre formula as
$$v_{2}(m!)=\sum_{i=1}^{\infty}\left\lfloor\frac{m}{2^{i}}\right\rfloor$$
As usual, by omitting the floor functions,
$$v_{2}(m!)<\sum_{i=1}^{\infty} \frac{m}{2^{i}}=m .$$
Thus, $L_{n}=m!$ implies the inequality
$$\begin{equation*} \frac{n(n-1)}{2}<m . \tag{2} \end{equation*}$$
In order to obtain an opposite estimate, observe that
$$L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right) \cdots\left(2^{n}-2^{n-1}\right)<\left(2^{n}\right)^{n}=2^{n^{2}} .$$
We claim that
$$\begin{equation*} 2^{n^{2}}<\left(\frac{n(n-1)}{2}\right)!\text{ for } n \geqslant 6 . \tag{3} \end{equation*}$$
For $n=6$ the estimate (3) is true because $2^{6^{2}}<6.9 \cdot 10^{10}$ and $\left(\frac{n(n-1)}{2}\right)!=15!>1.3 \cdot 10^{12}$.
For $n \geqslant 7$ we prove (3) by the following inequalities:
$$\begin{aligned} \left(\frac{n(n-1)}{2}\right)! & =15!\cdot 16 \cdot 17 \cdots \frac{n(n-1)}{2}>2^{36} \cdot 16^{\frac{n(n-1)}{2}-15} \\ & =2^{2 n(n-1)-24}=2^{n^{2}} \cdot 2^{n(n-2)-24}>2^{n^{2}} . \end{aligned}$$
Putting together (2) and (3), for $n \geqslant 6$ we get a contradiction, since
$$L_{n}<2^{n^{2}}<\left(\frac{n(n-1)}{2}\right)!<m!=L_{n}$$
Hence $n \geqslant 6$ is not possible.
Checking manually the cases $n \leqslant 5$ we find
$$\begin{gathered} L_{1}=1=1!, \quad L_{2}=6=3!, \quad 5!<L_{3}=168<6!, \\ 7!<L_{4}=20160<8!\quad \text{ and } \quad 10!<L_{5}=9999360<11!. \end{gathered}$$
So, there are two solutions:
$$(m, n) \in\{(1,1),(3,2)\} .$$

Like in the previous solution, the cases $n=1,2,3,4$ are checked manually. We will exclude $n \geqslant 5$ by considering the exponents of 3 and 31 in (1).
For odd primes $p$ and distinct integers $a, b$, coprime to $p$, with $p \mid a-b$, the Lifting The Exponent lemma asserts that
$$v_{p}\left(a^{k}-b^{k}\right)=v_{p}(a-b)+v_{p}(k) \text{.}$$
Notice that 3 divides $2^{k}-1$ if only if $k$ is even; moreover, by the Lifting The Exponent lemma we have
$$v_{3}\left(2^{2 k}-1\right)=v_{3}\left(4^{k}-1\right)=1+v_{3}(k)=v_{3}(3 k)$$
Hence,
$$v_{3}\left(L_{n}\right)=\sum_{2 k \leqslant n} v_{3}\left(4^{k}-1\right)=\sum_{k \leqslant\left\lfloor\frac{n}{2}\right\rfloor} v_{3}(3 k) .$$
Notice that the last expression is precisely the exponent of 3 in the prime factorisation of $\left(3\left\lfloor\frac{n}{2}\right\rfloor\right)!$. Therefore
$$\begin{align*} v_{3}(m!)=v_{3}\left(L_{n}\right) & =v_{3}\left(\left(3\left\lfloor\frac{n}{2}\right\rfloor\right)!\right) \\ 3\left\lfloor\frac{n}{2}\right\rfloor & \leqslant m \leqslant 3\left\lfloor\frac{n}{2}\right\rfloor+2 \tag{4} \end{align*}$$
Suppose that $n \geqslant 5$. Note that every fifth factor in $L_{n}$ is divisible by $31=2^{5}-1$, and hence we have $v_{31}\left(L_{n}\right) \geqslant\left\lfloor\frac{n}{5}\right\rfloor$. Then
$$\begin{equation*} \frac{n}{10} \leqslant\left\lfloor\frac{n}{5}\right\rfloor \leqslant v_{31}\left(L_{n}\right)=v_{31}(m!)=\sum_{k=1}^{\infty}\left\lfloor\frac{m}{31^{k}}\right\rfloor<\sum_{k=1}^{\infty} \frac{m}{31^{k}}=\frac{m}{30} \tag{5} \end{equation*}$$
By combining (4) and (5),
$$3 n<m \leqslant \frac{3 n}{2}+2$$
so $n<\frac{4}{3}$ which is inconsistent with the inequality $n \geqslant 5$.