jmc

All factors of $2^5\cdot 3^6$ that are perfect squares must be in the form $(2^m\cdot 3^n{)}^2=2^{2m}\cdot 3^{2n}$, where $0\le 2m\le 5$ and $0\le 2n\le 6$ for integers $m$ and $n$. Thus, $0\le m\le 2$ and $0\le n\le 3$, for a total of $3\cdot 4=\boxed{12}$ factors that are perfect squares.