Solution:

A difference of powers of $2$ is $2^a - 2^b = 2^b(2^c - 1)$ where $c = a - b \ge 1.$ Since $2^c - 1$ is odd, the odd part of the number determines $c$ and the power of $2$ determines $b,$ so distinct pairs $(b, c)$ yield distinct integers. It suffices to count pairs with $2^b(2^c - 1) \lt 1000.$

For $c = 1, 2, \ldots, 9$ the factor $2^c - 1$ is $1, 3, 7, 15, 31, 63, 127, 255, 511,$ and the number of allowed values of $b$ is $10, 9, 8, 7, 6, 4, 3, 2, 1$ respectively (the count for $63$ drops to $4$ because $63 \cdot 16 = 1008 \gt 1000,$ while $31 \cdot 32 = 992$ still fits).

The total is $10 + 9 + 8 + 7 + 6 + 4 + 3 + 2 + 1 = 50.$