One needn't memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\ d_1\ d_0 \ $ is $\rm\: (d_2\cdot 10 + d_1)\ 10 + d_0\ \equiv\ (d_2\cdot 3 + d_1)\ 3 + d_0\ (mod\ 7)\ $ since $\rm\ 10\equiv 3\ (mod\ 7)\:.\:$ So we compute the remainder $\rm\ (mod\ 7)\ $ as follows. Start with the leading digit then repeatedly apply the operation: multiply by $3$ then add the next digit, doing all of the arithmetic $\rm\:(mod\ 7)\:.\:$
For example, let's use this algorithm to reduce $\rm\ 43211\ \:(mod\ 7)\:.\:$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ d_n\ d_{n-1}\ $ by $\rm\ d_n\cdot 3 + d_{n-1}\:\ (mod\ 7),\:$ namely
$\rm\qquad\phantom{\equiv} \color{red}{4\ 3}\ 2\ 1\ 1$
$\rm\qquad\equiv\phantom{4} \color{green}{1\ 2}\ 1\ 1\quad $ by $\rm\quad \color{red}4\cdot 3 + \color{red}3\ \equiv\ \color{green}1 $
$\rm\qquad\equiv\phantom{4\ 3} \color{royalblue}{5\ 1}\ 1\quad $ by $\rm\quad \color{green}1\cdot 3 + \color{green}2\ \equiv\ \color{royalblue}5 $
$\rm\qquad\equiv\phantom{4\ 3\ 5} \color{brown}{2\ 1}\quad $ by $\rm\quad \color{royalblue}5\cdot 3 + \color{royalblue}1\ \equiv\ \color{brown}2 $
$\rm\qquad\equiv\phantom{4\ 3\ 5\ 2} 0\quad $ by $\rm\quad \color{brown}2\cdot 3 + \color{brown}1\ \equiv\ 0 $
Hence $\rm\ 43211\equiv 0\:\ (mod\ 7)\:,\:$ indeed $\rm\ 43211 = 7\cdot 6173\:.\:$ Generally the modular arithmetic is simpler if one uses a balanced system of representatives, e.g. $\rm\: \pm\{0,1,2,3\}\ \:(mod\ 7)\:.$ Notice that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9\:$).
Suppose $$2^n = \sum_{i=0}^{k-1} (m+i) = km + \frac{k(k-1)}{2}$$ for some $k>1$ and $m>0$, then $2^{n+1}=2km+k(k-1)=k(2m+k-1)$. Note then that this implies that $k=2^\ell$ for some $\ell > 0$, and therefore that $2m+k-1$ is odd, and therefore that in fact, $k=2^{n+1}$, and $2m+k-1=1$. Therefore we have $2m+2^{n+1}=2$, and $m+2^n = 1$, so $m=1-2^n\le 0$. Thus $2^n$ is not expressible as a sum of consecutive positive integers.
Edit:
I realized I posted this answer without really addressing the actual question, although I sort of answered in the comments. Everything other than the argument to show that powers of $2$ can't be expressed looked ok to me. I didn't follow your argument for the powers of $2$, which is why I wrote this answer.
Best Answer
From Bezout's Lemma, note that since $\gcd(7,11) = 1$, which divides $100$, there exists $x,y \in \mathbb{Z}$ such that $7x+11y=100$.
A candidate solution is $(x,y) = (8,4)$.
The rest of the solution is given by $(x,y) = (8+11m,4-7m)$, where $m \in \mathbb{Z}$. Since we are looking for positive integers as solutions, we need $8+11m > 0$ and $4-7m>0$, which gives us $-\frac8{11}<m<\frac47$. This means the only value of $m$, when we restrict $x,y$ to positive integers is $m=0$, which gives us $(x,y) = (8,4)$ as the only solution in positive integers.
If you do not like to guess your candidate solution, a more algorithmic procedure is using Euclid' algorithm to obtain solution to $7a+11b=1$, which is as follows.
We have \begin{align} 11 & = 7 \cdot (1) + 4 \implies 4 = 11 - 7 \cdot (1)\\ 7 & = 4 \cdot (1) + 3 \implies 3 = 7 - 4 \cdot (1) \implies 3 = 7 - (11-7\cdot (1))\cdot (1) = 2\cdot 7 - 11\\ 4 & = 3 \cdot (1) + 1 \implies 1 = 4 - 3 \cdot (1) \implies 1 = (11-7 \cdot(1)) - (2\cdot 7 - 11) \cdot 1 = 11 \cdot 2-7 \cdot 3 \end{align} This means the solution to $7a+11b=1$ using Euclid' algorithm is $(-3,2)$. Hence, the candidate solution $7x+11y=100$ is $(-300,200)$. Now all possible solutions are given by $(x,y) = (-300+11n,200-7n)$. Since we need $x$ and $y$ to be positive, we need $-300+11n > 0$ and $200-7n > 0$, which gives us $$\dfrac{300}{11} < n < \dfrac{200}7 \implies 27 \dfrac3{11} < n < 28 \dfrac47$$ The only integer in this range is $n=28$, which again gives $(x,y) = (8,4)$.