To determine $(n-1)$-tuples of nonnegative integers $a_1, a_2, \ldots, a_{n-1}$ such that

$$\left\lfloor \frac{m}{2^n - 1} \right\rfloor + \left\lfloor \frac{2m + a_1}{2^n - 1} \right\rfloor + \left\lfloor \frac{2^2m + a_2}{2^n - 1} \right\rfloor + \ldots + \left\lfloor \frac{2^{n-1}m + a_{n-1}}{2^n - 1} \right\rfloor = m$$

holds for all $m \in \mathbb{Z}$, we follow the below steps:

1. Rewriting the Floor Function Terms:
Each term in the sum involves a floor function $\left\lfloor \frac{2^k m + a_k}{2^n - 1} \right\rfloor$. For this entire sum to simplify to exactly $m$ for any integer $m$, the fractional parts must somehow balance out such that overall, we can reconstruct a precise integer result, i.e., bias the floors where needed.

2. Equate Sums and Analyze:
Let us start from the algebraic manipulation:

$$m = \left\lfloor \frac{m}{2^n - 1} \right\rfloor + \sum_{k=1}^{n-1} \left\lfloor \frac{2^k m + a_k}{2^n - 1} \right\rfloor$$

when rewritten implies:

$$\sum_{k=0}^{n-1} \left\lfloor \frac{2^k m + a_k}{2^n - 1} \right\rfloor \approx m \frac{2^n - 1}{2^n - 1}$$

3. **Determine Specific Values for $a_k$'s**:

As analyzing and checking multiple $m$ is not trivial without testing boundaries:

- Consider explicitly $a_k = k(2^n - 1) - (2^k - 1)m$.

Given this choice, compute each step:

$$a_k = (0)(2^n - 1) - (2^0 - 1)m = 0$$

$$a_k = (1)(2^n - 1) - (2^1 - 1)m = 2^n - 1 - m$$

This pattern as it holds till $n-1$, confirms that:

$$a_k = k(2^n - 1) - (2^k - 1)m$$

Suitably provides non-negative $a_k$ satisfying the equation as built when tested via any:

$$(a_1, a_2, \ldots, a_{n-1}) = \left(1(2^n - 1) - (2^1 - 1)m, 2(2^n - 1) - (2^2 - 1)m, \ldots, (n-1)(2^n - 1) - (2^{n-1} - 1)m \right)$$

Thus, the solution to the given problem is:

$$\boxed{\left(a_1, a_2, \ldots, a_{n-1}\right) = \left(1(2^n - 1) - (2^1 - 1)m, 2(2^n - 1) - (2^2 - 1)m, \ldots, (n-1)(2^n - 1) - (2^{n-1} - 1)m \right)}$$