We look at $2^n$, where $n$ ranges over the non-negative integers.
The key is the fact that $2^6$ has remainder $1$ on division by $9$. Using congruence notation, we have $2^6\equiv 1\pmod{9}$. Let $n$ be any non-negative integer. We can express $n$ as $6q+r$, where $0\le r\le 5$ ($q$ stands for quotient, $r$ for remainder).
It follows that
$$2^n=2^{6q+r}=(2^6)^q 2^r\equiv (1)^q 2^r\equiv 2^r\pmod{9}.$$
So the remainder when you divide $2^n$ by $9$ depends only on $r$. For $r=0$, $1$, $2$, $3$, $4$, and $5$, these remainders are, as you observed, $1$, $2$, $4$, $8$, $7$, and $5$.
To connect this with sums of (decimal) digits, observe that a decimal number like $6852$ is just $(6)(10^3)+(8)(10^2)+(5)(10^1) +(2)(10^0)$. But for any non-negative integer $k$, we have $10^k\equiv 1\pmod{9}$. So $6852\equiv 6+8+5+2 \pmod{9}$. Thus the remainder when $6852$ is divided by $9$ is the same as the remainder when $6+8+5+2$ is divided by $9$. Asimilar remark holds for any non-negative integer expressed in decimal form.
Since remainders when $2^n$ is divided by $9$ cycle with period $6$, and the "casting out nines" process gives us these remainders, the pattern you observed continues forever.
You extended the pattern to negative exponents. This is an interesting observation that I do not recall seeing before. Express $2^{-n}$ as a decimal, by noting that
$$2^{-n}=\frac{1}{2^n}=\frac{5^n}{10^n}.$$
Then (essentially) you looked at the digit sum (modulo $9$) of $5^n$. Modulo $9$, the numbers $5^n$ cycle with period $6$, for the same reason as with $2^n$.
Now calculate $5^n$ modulo $9$, for $n=0,1, 2, 3, 4, 5$. We get that $5^n$ is congruent in turn to $1$, $5$, $7$, $8$, $4$, and $2$ modulo $9$. So the pattern does indeed continue "backwards."
Remark: For a more general approach, please see Euler's Theorem.
You made a slight (but crucial) mistake. The correct formula should be
$$n = \frac{2^{x+2}-(4q+3^{\color{red}b}n)}{3^{b+1}}$$
And then it is indeed a trivial conclusion as
\begin{align*}
n&=\frac{2^{x+2}-(4q+3^bn)}{3^{b+1}}\\
n&=4\frac{2^x-q}{3^{b+1}}-\frac n3\\
3n&=4\frac{2^x-q}{3^b}-n\\
4n&=4\frac{2^x-q}{3^b}\\
\end{align*}
$$\therefore~n=\frac{2^x-q}{3^b}$$
And the last line is just the original assumption.
Best Answer
The number of digits of a number $n$ is $1+\lfloor \log_{10}n \rfloor$, so the number of digits of $2^k=1+\lfloor \log_{10} 2^k\rfloor=1+\lfloor k\log_{10} 2\rfloor$. You can use $\log_{10}2 \approx 0.30103000$ (or get more digits here) to justify your values.