By PIE, $|A|+|B|-|A \cap B| = |A \cup B|$. Substituting into the equation and factoring, we get that $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$, so therefore $A \subseteq B$ or $B \subseteq A$. WLOG $A\subseteq B$, then for each element there are $3$ possibilities, either it is in both $A$ and $B$, it is in $B$ but not $A$, or it is in neither $A$ nor $B$. This gives us $3^{5}$ possibilities, and we multiply by $2$ since it could have also been the other way around. Now we need to subtract the overlaps where $A=B$, and this case has $2^{5}=32$ ways that could happen. It is $32$ because each number could be in the subset or it could not be in the subset. So the final answer is $2\cdot 3^5 - 2^5 = \boxed{454}$.
~math31415926535
For any finite set $S$, let $|S|$ denote the number of elements in $S$. Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]