Solution:

Let $n$ be a zesty two-digit number, and let $x$ and $y$ be as in the problem statement. Clearly if both $x$ and $y$ are one-digit numbers, then $s(x) s(y) = n \neq s(n)$. Thus either $x$ is a two-digit number or $y$ is. Assume without loss of generality that it is $x$. If $x = 10a + b$, $1 \leq a \leq 9$ and $0 \leq b \leq 9$, then $n = 10a y + b y$. If both $a y$ and $b y$ are less than $10$, then $s(n) = a y + b y$, but if either is at least $10$, then $s(n) < a y + b y$. It follows that the two digits of $n$ share a common factor greater than $1$, namely $y$. It is now easy to count the zesty two-digit numbers by first digit starting with $2$; there are a total of $5 + 4 + 5 + 2 + 7 + 2 + 5 + 4 = 34$.