Call an integer 'n' that is not a square or a prime power or a square-free a 'square-in'.Let n be square-in. Then between n and (2 n) is there another square-in? This is a kind of 'variation' on Bertrand's Postulate.
For $n > 8$ the answer is certainly, as we just need to look at multiples of four, and at least one of them is not a square.
$n=1$ false, $n = 2$ false, $n = 3$ false, $n = 4$ true if we allow the upper bound of the range. $5 \leq n \leq 8$ all true due to either $8,12$. So depending on the definition, take $n \geq 4$ or $n > 4$.
EDIT: It has been noted that you require $n$ to be square-in, the first of which is $8$. So the claim is true.