For Q2, if $p \equiv 1 \bmod 3$, then $p$ may be written as $p=x^2+xy+y^2$ for some $x,y\in\mathbb{Z}$. Additionally if $b^2-b+1=p^m$ for some integer $m>0$, then it may be seen that $b \not\equiv 1 \bmod p$ and hence $\{b-1\}_p=b-1$. The above gives $(x^2+xy+y^2)^m=b^2-b+1$ or
$$
((x-y\omega)(x-y\omega^2))^m=(b+\omega)(b+\omega^2)
$$
where $\omega=exp(2 \pi i/3)$. For example, in your example,
$$
7=2^2+2+1
$$
and
$$
7^3=19^2-19+1
$$
as
$(2-\omega)^3$ times a unit of $\mathbb{Z}[\omega]$ is equal to $19+\omega$.
Also, for Q1, a partial answer (that can probably be extended further) is:
Let $1\leq a_1 < a_2 < \dots < a_n$ and let
$$
b=1+\sum_{i=1}^n 2^{a_i}.
$$
If $a_{n-1} < a_n - 2$, then $b-2^{a_n} < 2^{a_{n-1}+1} \leq 2^{a_n-2}$ and $b^2 - 2^{2a_n} < 2^{2a_n}$. Thus if we write $b^2$ as
$$
b^2 = 2^{2a_n} + \sum_{i \in A}2^i,
$$
the largest $i$ in $A$ satisfies $a_n+a_{n-1}+1 \leq i < 2a_n$. However, for such $b$, we would have to have
$$
b^2 = 2^{2a_n} + \sum_{i=1}^n 2^{a_n-a_1}
$$
if $\{b^2 - \{b-1\}_2\}_2 = 1$. However, $a_n-a_1 < a_n+a_{n-1}+1$. Hence $b$ with $a_{n-1} < a_n - 2$ cannot satisfy $\{b^2 - \{b-1\}_2\}_2 = 1$.
Write
$$
b = 1+\sum_{i=1}^n 2^{a_i} = 2^{a_n} + a
$$
and assume that $a>0$. Then
$$
b^2 = (2^{a_n} + a)^2 \geq 2^{2a_n+1}
$$
if and only if
$$
a^2 + 2^{a_n+1}a - 2^{2a_n} \geq 0.
$$
As $a \geq 0$, that means that
$$
b^2 \geq 2^{2a_n+1} \Leftrightarrow a \geq 2^{a_n}(\sqrt{2}-1).
$$
If $\{b^2 - \{b-1\}_2\}_2 =1$, then
$$
b^2 = 2^q + \sum_{i=1}^{n}2^{a_i-a_1}
$$
where
$$
q=\begin{cases}2a_n+1 \mbox{ if } a \geq 2^{a_n}(\sqrt{2}-1)\\2a_n\mbox{ otherwise.}\end{cases}
$$
It is the case that
$$
\{b-1\}_2 = \sum_{i=1}^n 2^{a_i-a_1} < 2^{a_n}.
$$
If $a < 2^{a_n}(\sqrt{2}-1)$ then $$b^2 - 2^{2a_n} = 2^{a_n+1}a + a^2 > 2^{a_n}.$$ Therefore if $a < 2^{a_n}(\sqrt{2}-1)$, we cannot have $\{b^2 - \{b-1\}_2\}_2 =1$. Also, if $a \geq 2^{a_n-1}$, then
$$
b^2 - 2^{2a_n+1} \geq 2^{2a_n-2}.
$$
$2^{2a_n-2} > 2^{a_n}$ if $a_n > 2$. Therefore if either $a < 2^{a_n}(\sqrt{2}-1)$ or ($a \geq 2^{a_n-1}$ and $a_n > 2$), we cannot have $\{b^2 - \{b-1\}_2\}_2 =1$.
Using this approach, it is possible to obtain:
Writing $b = 2^{a_n} + a$, with $0 < a < 2^{a_n}$, then $\{b^2 - \{b-1\}_2\}_2 = 1$ cannot happen when $a \in \mathbb{Z}$ and
$$
\frac{a}{2^{a_n}} \in (0,\sqrt{2}-1) \cup [\sqrt{2+\frac{1}{2^{a_n}}}-1,1).
$$
The length of the interval $[2^{a_n}(\sqrt{2}-1),2^{a_n}(\sqrt{2+\frac{1}{2^{a_n}}}-1))$ converges to $\frac{1}{2\sqrt{2}}$ as $a_n \rightarrow \infty$ and hence can contain at most one integer for each $a_n \geq 1$. Hence for each $a_n \geq 1$, there is at most one integer $a$ with $0 < a < 2^{a_n}$ such that $b = 2^{a_n} + a$ could possibly satisfy $\{b^2 - \{b-1\}_2\}_2 = 1$.
