Solution:

Here is a strategy requiring at most 18 drops. Drop the first phone from the 18th floor. If it breaks, drop the second phone from floors $1, 2, \ldots, 17$ in that order to determine the minimum breaking floor. Otherwise, drop the first phone from the $18+17=35$th floor. If it breaks, use the 16 remaining drops to test the second phone on floors 19 through 34. Otherwise, drop the first phone from the $18+17+16$th floor, and so on. If the first phone makes it to the $18+17+\cdots+5=161$st floor, then there are 4 drops left, more than enough to test it on the last two floors.

Suppose that there existed a strategy requiring at most 17 drops. There are 164 possible "strengths" of the Mithril II (the breaking floor could be $1, 2, \ldots, 163$ or undefined). When the strategy is applied to one of these strengths, the result is a sequence of intact (I) or broken (B) outcomes containing at most two B's and having length at most 17. In fact, we can arrange for the sequence to have length exactly 17 by declaring that, if at a given point the strategy stops and declares the phone strength, the outcomes of any unused drops are arbitrarily designated I. Then each phone strength has a "signature" of 17 letters, each B or I, with at most two B's. But the number of possible signatures is only
$$\binom{17}{0} + \binom{17}{1} + \binom{17}{2} = 1 + 17 + 136 = 154$$
This shows that there are some two phone strengths that have the same signature and therefore cannot be distinguished by the claimed strategy.