Finding one heavy bean among eight

Among $8$ beans that look the same, exactly one is a bit heavier than the rest. Using a two-pan balance, what is the fewest weighings that always suffices to identify it?

$2$. Weigh $3$ against $3$. If they balance, compare the remaining two. If not, take the heavier group of three and weigh two of them against each other. One weighing has only $3$ outcomes and cannot separate $8$ cases, so $2$ is optimal.