I know this can draw downvotes as it's not a complete solution or sounds more like an opinion poll (or perhaps this must have appeared elsewhere, in which case you're free to let me know), but I would like to seek your help regrading this topic.
I decided to continue working on Brocard's problem (i.e., the number of integer solutions to the Diophantine equation $n! + 1 = m^2$), and here's where I have reached:
The currently available solutions are as follows (along with the $m+1$ and $m-1$ values; format : $(n,m,m+1,m-1)$):
$(4,5,6,4)$
$(5,11,12,10)$
$(7,71,72,70)$
From these, we can see $2 \mid m + 1$ and $3 \mid m + 1$ and $m+1$ increases by a product of powers of $2$ and $3$.Hence, we can conjecture( just because we haven't found any more of solutions) that $m+1$ is always of the form $m+1$ = $2^x3^y$, and thus $m-1$ = $2^x3^y-2$. Thus, we get $$n! = 2^x3^y(2^x3^y -2) = 2^{x+1}3^y(2^{x-1}3^y-1)\space(\text{equivalent to to} 2(m+1)\times \frac{(m-1)}2)\longrightarrow(1)$$ I conjecture that if any other solution existed, it should be of this form.
Notice that $(2^{x-1}3^y-1 = \frac{m-1}2 \equiv 1 \pmod{2})\land(2^{x-1}3^y-1 = \frac{m-1}2 \equiv 2 \pmod{3})\forall x = y \neq 1$, so only powers of odd primes other than $3$ are present in its prime factoristaion for greater $n$. Thus we can group the numbers in the factorials as follows: $$4! = \color{red}{2^23}\times\color{blue}{2} = \color{red}{2(m+1)}\color{blue}{\frac{m-1}2}$$ $$5! = \color{red}{2^33}\times\color{blue}{5} = \color{red}{2(m+1)}\color{blue}{\frac{m-1}2}$$ $$7! = \color{red}{2^43^2}\times\color{blue}{5\times 7} = \color{red}{2(m+1)}\color{blue}{\frac{m-1}2}$$
[As of now the solutions to $x$ and $y$ for the available solutions are (format : $(n,x,y)$):
$(4,1,1)$
$(5,2,1)$
$(7,3,2)$
Apply these to the form at (1) and compare with the coloured ones]
And I believe that we have somehow grouped the numbers(reference:thread at Unsolved Problems group in groups.io, first comment by Kermit Rose; not sure if those who aren't signed in can access this) even if not so effectively (perhaps), and..... this is where I get stuck.
I am not able to proceed further from here as I am still at the basics. I was adviced against using the available solutions to arrive at a conclusion by Kermit Rose, as I may end up thinking that no more cases existed. But I believe the introduction of $x$ and $y$ into the equation must have opened up more possibilities of enquiry. Still I wonder why I can't proceed... a helping hand would indeed be appreciated.
PS: I doubt if it's the abc conjecture at work that limits the solutions for $n, x$ and $y$.
Updates:
https://groups.io/g/UnsolvedProblems/message/12809 - another observation by Kermit Rose. From his claim, I conjectured that for all solutions satisfy the property that if $k = \lfloor \frac{n+1}2 \rfloor, (k + 1 \mid m + 1)\lor(k-1 \mid m + 1), k -1 \neq 0$
http://unsolvedproblems.org/S73.pdf - an algorithmic approach to checking for solutions for the equation, by Robert D. Matson (from unsolvedproblems.org)
The problem is you don't have enough information as is ( which values could work for example). $$n!+1=m^2\implies n!=(m-1)(m+1)$$ But, that tells us is they are $n$-smooth ( whereas $m$ is $n$-rough) , not which factors go in either. We know exactly one $2$ goes in one of them. We can show by the original $$n!+1=m^2$$ that for example $$m^2\bmod 6=1,\forall n>2$$ which limits $m$ by showing it must be $\pm 1\bmod 6$ and so all $3$'s go to one side. We also can show $m\equiv \pm \bmod 5 \forall n>4$ . Unfortunately this doesn't really help much( in fact I bet my previous comments help as much). A better fact to use is that $\forall n>9, m\equiv 1,49,51,99\pmod {100}$
Here's the thing, $m$ is the variable we can easily say something about ( especially meaningfully). $n$ only really has the following ( to my knowledge). If you take $n$ to be an odd square, only $1$ of $n-1,n$ , can work. If $$(n-1)!+1=m_1^2$$ then $$n!+n=(\sqrt{n}m_1)^2$$ but then $$n!+1=m_2^2$$ implies that $$m_2^2-(\sqrt{n}m_1)^2<n$$ Further, the difference of squares is even, and therefore the jump between consecutive squares is under half of $n$ . This implies the smaller base is less than 1 quarter of $n$ . But $m$ grows faster than $n$ . Contradiction. But that saves no work unless you find a new $n$
And there are values of $n$ that have simple checks :
$$p=n+1$$ with $p$ prime has to have $p$ a Wilson prime. $$p=n+2$$ needs $p\equiv \pm 1\pmod 8$ by quadratic residues.
There are probably others.