Solution:

A positive integer has no $0$ in base three exactly when every digit is $1$ or $2.$ For $k = 1, 2, \ldots, 6$ there are $2^k$ such $k$-digit numbers, and all of them are at most $222222_3 = 728 \lt 2017.$

Since $2017 = 2202201_3,$ a seven-digit string of $1$s and $2$s is at most $2017$ exactly when it begins with $11,$ $12,$ or $21:$ any string beginning $22$ already beats $2202201_3$ at the third digit, since its digits are nonzero. That gives $3 \cdot 2^5 = 96$ seven-digit numbers.

The total is $2 + 4 + 8 + 16 + 32 + 64 + 96 = 222.$