Answer: 243

Small Hint:

Try two-subset partitions that keep small numbers and their pairwise products apart, such as separating $\{3, \ldots, 8\}$ from $\{9, \ldots, 80\}$

Big Hint:

For the upper bound, follow the powers of $3:$ once $3$ and $9$ are forced into different subsets, place $27$ and $81,$ and one more power creates a contradiction

Solution:

First, $m = 243$ works. Suppose $S = \{3, 4, \ldots, 243\}$ were partitioned into $T$ and $U$ with neither containing a product, and say $3 \in T.$ Then $9 = 3 \cdot 3$ must lie in $U,$ so $81 = 9 \cdot 9$ must lie in $T,$ and then $243 = 3 \cdot 81$ must lie in $U.$ Now consider $27:$ if $27 \in T,$ then $3 \cdot 27 = 81$ puts a product in $T;$ if $27 \in U,$ then $9 \cdot 27 = 243$ puts one in $U.$ Either way we reach a contradiction.

For $m = 242,$ the partition $T = \{3, \ldots, 8\} \cup \{81, \ldots, 242\}$ and $U = \{9, \ldots, 80\}$ avoids products: two elements of $\{3, \ldots, 8\}$ multiply to something in $[9, 64] \subseteq U,$ any product involving an element of $\{81, \ldots, 242\}$ is at least $3 \cdot 81 = 243 \gt 242,$ and two elements of $U$ multiply to at least $81 \gt 80.$

Hence the smallest such $m$ is $243.$