Generalising an answer to $a!=7!5!3!$, consider the order of magnitude of the left and right sides. Factorials can be approximated well using Stirling's approximation but even crude bounds help. Factorials grow very quickly, so (at least past, say, $3!=6$) two factorials are either "exactly the same" or "very far apart". If $a!=b!5!3!$ then the right-hand side will be dominated by the largest factorial, $\max\{b!,5!\}$, but $b!5!3!$ won't be "very much" bigger than $\max \{b!,5! \}$ so it seems $a$ cannot be "much" larger than $\max \{ b,5 \}$. (In fact no non-trivial solutions are known for any equation $a!=b!\times c! \times \cdots$ where, without loss of generality, $b \ge c \ge \cdots$, for which $a > b + 3$.)
Let's write $a=b+k$ and evaluate the known factorials so $(b+k)!=720b!$; clearly $k \ge 1$ and our hope (justified by handwaving above) is that $k$ can't be much larger than one, so we should have only a few cases to check. We need to solve
$$(b+1)(b+2)\cdots (b+k) = 720 $$
so an obvious inequality is
$$ (b+1)^k \le 720 \le (b+k)^k $$
and if $k>1$ these inequalities become strict. It would be useful to dispose of $k=1$ for this reason: when $k=1$ the above collapses to $b+1=720$ so we obtain $b=719$, $a=b+k=720$, and $720!=719!5!3!$ is a solution.
We could also make a better estimate using the arithmetic mean of $b+1, b+2, \dots, b+k$ which is $b + (k+1)/2$, so that $(b + (k+1)/2)^k \approx 720$. The AM-GM inequality then gives that $(b + (k+1)/2)^k \ge (b+1)\cdots (b+k)$, and since $k>1$ we can make this inequality strict. We now have an improved upper bound, and taking the $k$-th root obtain
$$ b+1 < \sqrt[k]{720} < b + \frac{k+1}{2} \implies \sqrt[k]{720} - \frac{k+1}{2} < b < \sqrt[k]{720} - 1$$
For each $k$ we can see which values of $b$ are plausible, then check if any give $(b+1)\cdots (b+k)=720$.
For $k=2$, $\sqrt{720} = 26.83\dots$ gives $25.33\dots < b < 25.83\dots$ with no integer solution.
For $k=3$, $\sqrt[3]{720} = 8.96\dots$ gives $6.96\dots < b < 7.96\dots$ and hence $b = 7$, $a=b+k=10$. Indeed $8 \times 9 \times 10 = 720$, so $10! = 7! \times 5! \times 3!$ is a solution.
For $k=4$, $\sqrt[4]{720} = 5.18\dots$ gives $2.68\dots < b < 4.18\dots$ so $b=3$ or $b=4$. But neither work, since $4 \times 5 \times 6 \times 7 > 720$ and so clearly $5 \times 6 \times 7 \times 8 > 720$ also.
For $k=5$, $\sqrt[5]{720} = 3.72\dots$ gives $0.72\dots < b < 2.72\dots$ so $b=1$ or $b=2$. For $b = 1$, we need to check $2 \times 3 \times 4 \times 5 \times 6 = 720$, which it does. So we have found the solution $6! = 1! \times 5! \times 3!$. For $b = 2$, clearly the product $3 \times 4 \times 5 \times 6 \times 7 > 720$.
For $k=6$, $\sqrt[6]{720} = 2.99\dots$ gives $-0.51 \dots < b < 1.99 \dots$ and moreover for $k>6$, this means $\sqrt[k]{720} < 3$ and $b < 2$. So it only remains to check $b=0$ and $b=1$ for $k \ge 6$. Either way, $b! = 1$ so our original equation becomes $a!=720$. Since $a=b+k \ge 6$, we check if $6!=720$. It does, so $6! = 0! \times 5! \times 3!$ is a solution. But we can't have a solution for $b=1, k\ge 6$ or $b=0, k\ge 7$ or else $a!=(b+k)!\ge 7!>720$.
Overall we obtain the exact solutions $720! = 719! \times 5! \times 3!$, $10! = 7! \times 5! \times 3!$, $6! = 1! \times 5! \times 3!$ and $6! = 0! \times 5! \times 3!$. Examining our results for $k=2$ and $k=4$ more closely, our method has also found the "near(ish) misses":
$$\frac{27!}{25!5!3!} = 0.975; \frac{28!}{26!5!3!} = 1.05; \frac{6!}{5!3!2!} = 0.5; \frac{7!}{5!3!3!} = 1.1\dot{6} $$
Note that using the AM-GM inequality for $(b+1)(b+2)\cdots (b+k)$ ended up giving us an improved lower bound for $b$, but this didn't turn out to be essential. In the end the upper bound
$b < \sqrt[k]{720} - 1$ was more important, since $\log_3 720 = 5.98\dots < 6$ (i.e. $720 < 729 = 3^6$) meant that for $k \ge 6$ we have $b<2$ so $b \in \{0,1\}$. This is why our procedure terminates at the sixth step: once we checked products of at least six consecutive integers starting at $(b+1)\in \{1,2\}$ we knew we had found all solutions. It's clear this procedure will work for all questions of the form $a! = b! \times n$ where $n$ is a given integer, and take $\lceil \log_3{n} \rceil$ steps (where $\lceil \dots \rceil$ is the ceiling function i.e. round the log up). Had we persisted with our cruder lower bound $\sqrt[k]{720} - k < b$, we'd have terminated at the same stage but needed to check a few more cases manually along the way (e.g. for $k=2$ we'd have had $24.83\dots < b < 25.83\dots$ and needed to see if $b=25$ works), which, while inconvenient, would not be fatal to our endeavour.
For $a! = b! \times n$ we see we need to check one case when $k=1$ ($a=n$, $b=n-1$, which always works); for $2 \le k \le \lfloor \log_3{n} \rfloor$ and using the stricter bounds, $b$ lies in the interval $(\sqrt[k]{n} - (k+1)/2, \sqrt[k]{n} - 1)$ which is open with width $(k-1)/2$ so contains at most $\lfloor k/2 \rfloor$ integer values to check; for $k = \lceil \log_3{n} \rceil$ we need to check two cases ($b=0$ and $b=1$, although their details will be identical). So in the worst case the number of possible solutions this procedure needs to check is
$$1 + (1 + 1) + (2 + 2) + \dots + (\lfloor \frac{\lfloor \log_3{n} \rfloor}{2} \rfloor + \lfloor \frac{\lfloor \log_3{n} \rfloor}{2} \rfloor) + 2$$
which is $\mathcal{O}\left((\log_3{n})^2\right)$.
