Let $z$ be an integer $> 1$ and let $M$ be the set of all numbers of the form $z_k = 1+z + \cdots+ z^k, \  k = 0, 1,\ldots$. Determine the set $T$ of divisors of at least one of the numbers $z_k$ from $M.$
To determine the set \( T \) of divisors of at least one of the numbers \( z_k \) from \( M \), we need to analyze the properties of the numbers \( z_k = 1 + z + \cdots + z^k \) for \( k = 0, 1, \ldots \).

1. **Consider the case when \(\gcd(n, z) > 1\):**
   - If \( \gcd(n, z) > 1 \), then \( n \) shares a common factor with \( z \).
   - Notice that \( z_k = 1 + z + z^2 + \cdots + z^k \) can be written as:
     \[
     z_k = \frac{z^{k+1} - 1}{z - 1} \quad \text{for} \quad z \neq 1
     \]
   - Since \( \gcd(n, z) > 1 \), \( n \) divides \( z \), and thus \( n \) cannot divide \( z_k \) because \( n \) would not divide \( 1 \) (the constant term in the numerator).

2. **Consider the case when \(\gcd(n, z) = 1\):**
   - If \( \gcd(n, z) = 1 \), then \( n \) and \( z \) are coprime.
   - Consider the \( n \) numbers \( z_0, z_1, \ldots, z_{n-1} \). If none of these numbers is divisible by \( n \), then by the pigeonhole principle, there must be two numbers \( z_i \) and \( z_j \) such that:
     \[
     z_i \equiv z_j \pmod{n} \quad \text{with} \quad 0 \leq i < j < n
     \]
   - This implies:
     \[
     z_j - z_i = z^{i+1} (1 + z + \cdots + z^{j-i-1}) \equiv 0 \pmod{n}
     \]
   - Since \( \gcd(n, z) = 1 \), \( n \) must divide \( 1 + z + \cdots + z^{j-i-1} \), which is one of the \( z_k \) for some \( k \).

Therefore, the set \( T \) of divisors of at least one of the numbers \( z_k \) from \( M \) consists of all integers \( n \) such that \( \gcd(n, z) = 1 \).

The final answer is \( \boxed{ T = \{ n \in \mathbb{Z}^+ \mid \gcd(n, z) = 1 \} } \).