Answer: $m=n^2+2n$. Proof: Obviously, $m\geq (n+1)^2$ is not possible, since two squares would be in the same subset and the product of two perfect squares is a perfect square. Now, the subsets $A_k=\{k^2,k^2+1,...,k^2+2k\}$ for $k=1,...,n$ form a partition of the set $\{1,...,m\}$. For $0\leq a<b\leq 2k$ we have \begin{align*} (k^2+a)(k^2+b) &= k^4+(a+b)k^2+ab < \left(k^2+\frac{a+b}{2}\right)^2 \text{ and } \\ \left(k^2+\frac{a+b-1}{2}\right)^2 &= k^4+(a+b-1)k^2+ab+\left(\frac{a-b}{2}\right)^2-(a+b)+\frac{1}{4} \\ &\leq k^4+(a+b-1)k^2+ab+\left(\frac{2k}{2}\right)^2-1+\frac{1}{4} \\ &< (k^2+a)(k^2+b), \end{align*}so $(k^2+a)(k^2+b)$ is not a perfect square. Hence, the product of two different elements of $A_k$ is not a perfect square.