General equations of factorials equal to the product of other factorials, e.g. $a! = b! \times c! \times d!$, have been asked before on this site and turn out to be an open problem, though only four non-trivial solutions to such equations are known.*
The very specific question $a! = 7! \times 5! \times 3!$ has also been asked here before. Despite being a straightforward question, in the sense that it is purely computational, it received answers using a range of different methods, and is still somewhat "interesting" because it relates to one of the few non-trivial solutions to the general case.
So, I would like to ask for all non-negative integer solutions to $a! = b! \times 5! \times 3!$. This seems a good "intermediate" question between the previous two: sufficiently more specific than the general case that I believe it can be solved, for example using methods that generalise the problem $a!=7!5!3!$, but rather more interesting than that specific case since it isn't purely computational and will include additional (trivial) solutions, as well as the non-trivial one! I will post an attempt at an answer, but the previous question suggests there will be a wide range of possible approaches so I would be very interested to see how other people tackle it.
(I'm hoping that seeing some methods for solving a more general problem like $a!=b!5!3!$ might at least hint at why finding non-trivial questions to such equations is so rare. A couple of other "intermediate" questions suggest themselves: $a!=7!b!c!$, $a!=5!b!c!$ and $10!=b!c!d!$ would be harder in the sense of solving for three variables, but each would catch two of the non-trivial solutions, while $a!=7!b!c!d!$ would catch three of the four. These might make good follow-up questions, unless someone posts an answer here that easily extends to handle all these cases, but aside from these it seems there are few other "factorial is a product of factorials" problems that would be suited for Math SE without being either too trivial or too difficult.)
$(*)$ According to this answer on MathOverflow, these will be the only non-trivial solutions if Baker's explicit abc conjecture is correct (see Nair and Shorey's 2016 paper in the Journal of Number Theory).
Best Answer
Your equation can be written as:
$$a! = 720 \times b!$$
Let's enumerate all the ways to write $720$ as the product of consecutive integers:
It turns out that no factor of 720 satisfies the equations $n(n+1) = 720$ or $n(n+1)(n+2)(n+3) = 720$, so there are no expansions with two or four factors. Nor can there be an expansion with 7 or more distinct factors, because the product would have to be at least $7! > 6! = 720$. But there are expansions with 1, 3, 5, and 6 factors.
We want both sides to be a factorial, so just set $b$ to be 1 less than the smallest factor of each expansion.
This gives us the solution set:
$$(a, b) \in \{ (720, 719), (10, 7), (6, 1), (6, 0) \}$$