We are solving $x! = y! (y+1)! = (y+1)(y!)^2$ over the positive integers.
We can prove the following. Given $x$, and $p$ the largest prime less than or equal to $x$, then $y=p-1$. In fact, if $y\geq p$, then $p^2 | y!(y+1)!$, but $p^2$ does not divide $x!$. If $y < p-1$, then $p$ does not divide $y! (y+1)!$, but $p | x!$.
Note the above is the case with $x=10$ and $y=6$.
My hope is that using some fact about prime numbers, such as the prime density theorem, we can prove that for $x \geq M$ for some $M$, there are no solutions. That is, there are a finite number of solutions.
EDIT: Using the generalized Bertrand's postulate, for large enough numbers, there is a prime $p$ with $x \geq p\geq 3/4 x$. Then if a solution exists,
$$x! = \Gamma(x+1) = \Gamma(p+1)\Gamma(p+2) \geq \Gamma(p+1)^2 \geq \Gamma(3x/4+1)^2.$$
Substituting Stirling's formula gives
$$(x/e)^x \sqrt{2 \pi x} \geq [(3x/4e)^{3x/4} \sqrt{2 \pi (3x/4)}]^2$$
which simplifies to
$$\sqrt{2 \pi x} \geq \frac{3 \pi}{2} x \left(\left(3/4e\right)^3 x \right)^{x/2}.$$
However, as $x \to \infty$, the RHS goes to infinity faster than the LHS. Hence, the inequality is violated and we conclude that there are a finite number of solutions. (All of the above can be made rigorous, knowing that Stirling's formula is an asymptotic result, and dealing with limits.)
Easiest way to estimate that number is to employ the Stirling approximation: you have
$$ n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$
(this is asymptotic notation), or even more roughly speaking, $ n! \approx e^{n\ln n} $ as order of magnitude. In your case, say $ n = 10^{80} $, you have
$$ n! \approx e^{80\cdot\ln{10}\cdot10^{80}} \approx 10^{10^{81.9}}. $$
(EDIT:
I had written 81 instead of 81.9 at the exponent; not only was the rounding incorrect, but especially, rounding exponents needs a lot of care when talking about orders of magnitude!)
By the way, this is also the plausible order of magnitude (in natural units) for the volume of the phase space of the observable universe.
Pictorially speaking, the decimal expansion of this number would need to convert every single particle in the observable universe into a digit in order to be written out.
For an even more pictorial comparison, the age of the universe is thought to be about $ 10^{17} $ seconds. This is far less than the number of digits you would need to write out the above number. Even if you could churn out, say, a billion of digits per seconds, you would still need about $ 10^{82}/10^9 = 10^{73} $ seconds to complete the task, which is a ridiculously prohibitory amount of time.
Best Answer
The factorial function is extended by the $\Gamma$ function. The relation is $$(n-1)! = \Gamma(n) = \int_0^\infty t^{n-1} e^{-t}\, dt$$ This can be analytically continued as a meromorphic function in the complex plane. Ref: John Conway's book on Complex Analysis.