Looks easy

The problem is equivalent to finding when $x!$ is going to be a square number. Bertrand's postulate is a theorem stating that for any integer $n>3$ , there always exists at least one prime number $p$ with

$n<p<2n-2.$ So,we have x! will have a prime with power one when $\frac{x}{2}>3$ that is $x>6.$ Hence by bertrand's postulate, we have for $x>6$ (roughly) there will be a prime which has exponent only one in $n!$ .So,we only need to consider small cases.

After working those small ones we get, $1$ is the only solution

We know that $x!$ will be a perfect square iff it can be decomposed to primes whose powers are all even. But, by Legendre's Theorem, we will find that there will be at least one prime whose power will be odd. And hence this completes the proof

Just searched the first 1000 integers using Mathematica

Select[Range@1000, IntegerQ[Sqrt[#!]] &]