How to solve $sqrt{x!y!}=xy$ for $(x,y)inmathbb{Z}_{geq0}timesmathbb{Z}_{geq0}$?

$$frac{(x-1)!}{x}=frac{y}{(y-1)!}.$$

The possible values of the $LHS$ are

$$1,frac12,frac23,frac32,frac{24}5cdots$$

There is no need to go further, as the next inverses are below $dfrac12$. The only solutions correspond to

$$1leftrightarrow1$$ and $$frac23leftrightarrow frac32.$$

Note that $(x-1)! > x$ for $x > 4$. So for $x,y > 4$ there are no solutions. Suppose without loss of generality that $x leq y$. Then we can check each possible value of $x$ separately.

If $x = 1$ then we get $(y-1)! = y$ which has solution $y=1$. If $x = 2$ then $(y-1)! = 2y$ which has no solutions. For $x = 3$, $2(y-1)! = 3y$ which gives $y = 4$. $x = 4$ also yields no solutions so we are left with the three solutions $(x,y) in {(1,1),(3,4),(4,3)}$

You can easily show that $i^2 / i! leq 2$ for $iinmathbb N$. So $x,y < 6$. From that is just a matter of checking $36$ pairs.

>>> for i in range(7):
...     f = math.factorial(i)
...     print(i, i*i, f, i*i/f)
... 
0 0 1 0.0
1 1 1 1.0
2 4 2 2.0
3 9 6 1.5
4 16 24 0.6666666666666666
5 25 120 0.20833333333333334
6 36 720 0.05

So the solutions are $(1,1), (3,4), (4,3)$.