When is $b^2 - \{b-1\}_2$ with odd $b$ a perfect power of $2$? (The bracket-notation explained below)

Question

For the complete extraction of the factor $p$ and its powers from a natural number $m$ let's define the notation $$ \{n\}_p := { n \over p^{\nu_p(n)}} $$
While looking at the question of existences of 2-step-cycles in the generalized Collatz-problem for $mx+1$ I arrived at the question for which odd $b$ the evaluation of $f(b)$ $$ f(b) = b^2 - \{b-1\}_2 \overset{?}=2^S \tag 1 $$ is a perfect power $ 2^S$.
I know, that $b=3$ and then $S=3$ is a solution (the famous $3^2-1=2^3$). Generating a list for sequential $b$ didn't suggest any useful pattern so far, so this didn't help me with a clue.

---

^{I've put the same question also at MathOverflow having now some helpful answers, but still no full solution}