If at most one term in a sequence $a$ is non-zero, it is immediate that it is $m$-powerful for all $m \in \mathbb{N}$. We will see that there are no other 30-powerful sequences (or indeed, $m$-powerful for any even $m$), but that there are many other sequences that are $m$-powerful for all odd $m$.

We begin our analysis with a simple lemma.

Lemma. For $m \in \mathbb{N}$, let $P_m(x, y) = (x+y)^m - x^m - y^m$, $x, y \in \mathbb{R}$. Then $P_m(x, y) = 0$ whenever $xy = 0$. Moreover, $P_m$ has no other roots if $m$ is even, but $P_m(x, -x) = 0$ if $m$ is odd.

Proof. Let $p(t) = (1+t)^m - 1 - t^m$, $t \in \mathbb{R}$. By expansion of $(1+t)^m$, we see that $p$ is a polynomial with non-negative coefficients, and so $p(t) > 0$ for $t > 0$. Suppose additionally that $m$ is even. If $-1 \le t < 0$, then $(1+t)^m - 1 \le 0$, so $p(t) \le -t^m < 0$. If $t < -1$, then $0 < (1+t)^m < t^m$, so $p(t) < -1 < 0$. We conclude that $0$ is the only root of $p$ when $m$ is even. If $xy \ne 0$, then $P_m(x, y) = x^m p(t)$, where $p$ is as above, and $t = y/x \ne 0$. It follows that $P_m(x, y) \ne 0$ if $m$ is even and $xy \ne 0$:

The statements that $P_m(x, y) = 0$ when $xy = 0$ (regardless of the parity of $m$) and $P_m(x, -x) = 0$ when $m$ is odd, both follow immediately. $\square$

Fix an arbitrary sequence $a = (a_k)$. For $m, n \in \mathbb{N}$, let
$$S(m, n) = \left( \sum_{k=1}^{n} a_k \right)^m - \sum_{k=1}^{n} a_k^m,$$
and let $D(m, n) = S(m, n+1) - S(m, n)$. The condition "a is m-powerful"
says that $S(m, n) = 0$, $n \in \mathbb{N}$, and so we also have $D(m, n) = 0$.

Define $x_n = a_{n+1}$ and $y_n = \sum_{k=1}^{n} a_k$. The equation $S(m, n) = 0$ can be written as
$$y_n^m = \sum_{k=1}^{n} a_k^m,$$
so if this equation holds, then the equation $D(m, n) = 0$ can be written as
$$(x_n + y_n)^m - x_n^m - y_n^m = 0.$$

Suppose now that $m$ is even and $a$ is $m$-powerful. By the lemma, we conclude that $x_n y_n = 0$ for all $n \in \mathbb{N}$. We will use these last equations to prove by induction on $i \in \mathbb{N}$, that at most one of the terms $a_1, ..., a_i$ is non-zero; we call this property $A_i$.

$A_1$ is trivially true, so suppose that $A_i$ is true for a specific $i = n \in \mathbb{N}$. If $a_1, \dots, a_n$ are all zero, then $A_{n+1}$ follows immediately, so we may assume that exactly one of these terms is non-zero. But now $y_n \ne 0$, so the equation $x_n y_n = 0$ implies that $x_n = a_{n+1} = 0$, and we again deduce $A_{n+1}$. This finishes the proof that if $m$ is even, then the $m$-powerful sequences are those with at most one non-zero term. Part (a) follows.

The analysis is similar for $m$ odd, but now $P_m$ has other roots in the lemma. By considering these roots, our analysis readily leads us to see that $a = ((-1)^n)_{n=1}^\infty$ is $m$-powerful for every $c \in \mathbb{R}$. Taking any non-zero $c$, we get an example with the properties required in (b).