In the prime factorization of the LHS, there are no primes greater than $2m-1$. By Bertrand's Postulate, there must be a prime strictly between $2m-1$ and $4m-2$. Thus if $4m-2<4m\leq\frac{m(m+1)}{2}$, then a prime will surely divide the factorial in the RHS but will not divide the LHS, e.g. if $4m\leq \frac{m(m+1)}{2}\Leftrightarrow m\geq 7$. We check $1\leq m\leq 6$ and see that $\boxed{m\in\left\{1,2,3,4\right\}}\mbox{.}$

Vworldv wrote: In the prime factorization of the LHS, there are no primes greater than $2m-1$. By Bertrand's Postulate, there must be a prime strictly between $2m-1$ and $4m-2$. Thus if $4m-2<4m\leq\frac{m(m+1)}{2}$, then a prime will surely divide the factorial in the RHS but will not divide the LHS, e.g. if $4m\leq \frac{m(m+1)}{2}\Leftrightarrow m\geq 7$. We check $1\leq m\leq 6$ and see that ${m\in\left\{1,2,3,4\right\}}\mbox{.}$ yes,my solution is exactly the same as yours. but before I worked it out,I had thought about using inequality for a long time.I think that the following ineq holds: when $m\ge 5$,$1!3!...(2m-1)!>(\frac{m(m+1)}{2})!$. can someone prove it?

By Bertrand $m\ge 7$ fails because RHS has a prime factor not included in the LHS. For $m=5,6$ the same is true, just not by bertrand. Then $\boxed{m=1,2,3,4}$ can be checked and confirmed as our answer