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 $(x,y) = (8,4)$.
Let $n=k^3.$ If $n$ has $6$ digits, the largest it could be is $(6\cdot 9)^3 = 157464$, which shows that one digit must be $1$. So in fact the largest it could be is $(1+5\cdot 9)^3 = 97336$ which is too small. So there are no $6$ digit (or larger) solutions. (I think this is the gist of @dezdichado's comment.)
Because the cube root of $100,000 = 46.4\ldots$, we know that $k\leq 46.$ Using your observation that $k\equiv -1, 0, 1 \pmod{9}$ leaves us with only these $16$ possibilities for $k$:
$$1, 8, 9, 10, 17, 18, 19, 26, 27, 28, 35, 36, 37, 44, 45, 46.$$
We can just check these or we can play a bit more:
Since $(5\cdot 9)^3 = 91125$, we see that a $5$-digit solution must have at least one non-$9$ digit. Since $(4\cdot 9+8)^3 = 85184$, the largest the first two digits can be is $79$, so at least one digit is at most $7$. Since $4\cdot 9 +7 = 43$ we can cross off the last three numbers from the list.
Of these, $1, 8, 17, 18, 26, 27$ work.
A couple of observations: It surprises me that the last two pairs are consecutive, and both are of the shape $-1, 0 \pmod{9}$ so I feel like I've missed something. Since $1$ is a solution, it doesn't seem like modular considerations would eliminate $1 \pmod{9}$.
Best Answer
Hint 1) $a$ must be a multiple of $5$.
Hint 2) You need only consider multiples of the $(3,4,5)$ and $(5,12,13)$ triangles.