jmc

We prime factor the given product as $2^{10}\cdot 3^2\cdot 5.$ Recall that a number is a perfect square if and only if all of its prime factors are raised to an even power; so $f$ is a factor if and only if $f=2^{2a}\cdot 3^{2b}$ for $0\le 2a\le 10$ and $0\le 2b\le 2.$ So we have $6$ choices for $a$ and $2$ choices for $b,$ leading to a total of $6\cdot 2=\boxed{12}$ possibilities.