Answer: 454

Solution:

Let $a = |A|,$ $b = |B|,$ and $i = |A \cap B|,$ so $|A \cup B| = a + b - i.$ The condition $ab = i(a + b - i)$ rearranges to $$\begin{aligned} &ab - ia - ib + i^2 \\ &= (a - i)(b - i) = 0, \end{aligned}$$ so $|A \cap B| = |A|$ or $|A \cap B| = |B|.$ Since $A \cap B$ is a subset of each, that means $A \subseteq B$ or $B \subseteq A.$

For pairs with $A \subseteq B,$ each of the $5$ elements independently lies in neither set, in $B$ only, or in both: $3^5 = 243$ pairs. Likewise $243$ pairs satisfy $B \subseteq A,$ and the pairs counted twice are exactly those with $A = B,$ of which there are $2^5 = 32.$ The answer is $243 + 243 - 32 = 454.$