Idea: show that $1!3!5!\cdots(2n-1)!\geqslant(4n-1)!$, so that $m\geq4n-1$. By Bertrands Postulate it will follow that there are no solutions because there is a prime between $2n-1$ and $m$.
It suffices that $1!3!5!\cdots(2n-3)!\geqslant2n(2n+1)\cdots(4n-1)$. Let $n\geq12$. We have
$$\begin{array}{c}(2n-3)!&=1\cdot&\cdots&n\cdot&n+1&\cdots&\cdot(2n-5)&\cdot(2n-4)&\cdot(2n-3)\\
&\geq&2^{n-2}&\cdot\frac{2n}2&\cdot\frac{2n+2}2&\cdots&\cdot\frac{4n-10}2&\cdot\frac{4n-8}2&\cdot\frac{4n-6}2\\
(2n-5)!&=1\cdot&\cdots&n\cdot&n+1&\cdots&\cdot(2n-5)\\
&\geq&2^{n-4}&&\cdot\frac{2n+1}2&\cdots&\cdot\frac{4n-11}2\\
\end{array}$$
so $(2n-5)!(2n-3)!\geq2n(2n+1)\cdots(4n-10)\cdot(4n-8)(4n-6)$. There are $7$ factors missing to get $2n(2n+1)\cdots(4n-1)$. We'll get these factors from the remaining factorials.
Since $x!\geq2\cdot3x\geq2(x+10)$ for $x\geq5$, we have
$$\begin{array}{l}&(2n-19)!&(2n-17)!&(2n-15)!&(2n-13)!&(2n-11)!&(2n-9)!&(2n-7)!\\
\geq&(4n-9)&\cdot(4n-7)&\cdot(4n-5)&\cdot(4n-4)&\cdot(4n-3)&\cdot(4n-2)&\cdot(4n-1)\end{array}$$
(note that $2n-19\geq5$ because $n\geq12$) so
$(2n-19)!\cdots(2n-3)!\geq2n(2n+1)\cdots(4n-1)$.
Checking $n<12$:
for $m>1$, $m!$ has a prime divisor with multiplicity $1$ so it suffices to check primes $2n-1$:
- $2n-1=19$: $23$ is prime, so $m\leq22$. But $5!7!9!>20\cdot21\cdot21$, so LHS>RHS.
- $2n-1=17$: $19$ is prime, so $m\leq18$. But $5!>18$, so LHS>RHS.
- $2n-1=13$: $17$ is prime, so $m\leq16$. But $5!7!9!>14\cdot15\cdot16$, so LHS>RHS.
- $2n-1=11$: $13$ is prime, so $m\leq12$. But $5!>12$, so LHS>RHS.
- $2n-1=7$: this works and gives $m=10$, $n=4$.
- $2n-1=5$: this works and gives $m=6$, $n=3$.
- $2n-1=3$: this works and gives $m=3$, $n=2$.