To prove that if the sum of all positive divisors of $n \in \mathbb{Z}^{+}$ is a power of two, then the number of divisors of $n$ is also a power of two, we first introduce some notation and known results:

### Notation and Definitions

1. Let $n$ be a positive integer.
2. Denote the set of positive divisors of $n$ as $\{d_1, d_2, \ldots, d_k\}$, where $d_1 = 1$ and $d_k = n$.
3. The sum of all positive divisors of $n$ is given by:
$$\sigma(n) = \sum_{i=1}^{k} d_i.$$

### Known Facts

- Prime Power Divisors: If $n = p_1^{a_1} p_2^{a_2} \ldots p_m^{a_m}$ is the prime factorization of $n$, then the number of positive divisors of $n$ is:
$$k = (a_1 + 1)(a_2 + 1) \ldots (a_m + 1).$$
- Sum of Divisors: The sum of the divisors for $n$ with the same prime factorization is:
$$\sigma(n) = (1 + p_1 + p_1^2 + \ldots + p_1^{a_1})(1 + p_2 + p_2^2 + \ldots + p_2^{a_2}) \ldots (1 + p_m + p_m^2 + \ldots + p_m^{a_m}).$$

### Proof

Given that $\sigma(n)$ is a power of two, let's denote it by $2^b$, where $b$ is a non-negative integer.

For $\sigma(n)$ to be a power of two, each factor $(1 + p_i + p_i^2 + \ldots + p_i^{a_i})$ must also be a power of two because if one factor is not a power of two, $\sigma(n)$ cannot be a power of two.

Let’s consider each factor of $\sigma(n)$:

- For a prime $p_i$, $1 + p_i + p_i^2 + \ldots + p_i^{a_i}$ being a power of two implies that the sequence $(1, p_i, p_i^2, \ldots, p_i^{a_i})$ must sum to a power of two.

- This can happen when $a_i = 1$; and therefore, $n$ is composed of distinct prime powers such as $n = 2^1, 3^1$, etc.

Thus, the structure $n$ should support $k$ being a power of two to ensure $\sigma(n)$ is a power of two:

1. If each $(a_i + 1)$ for the corresponding prime $p_i$ is a power of two, then their product, $k$, the number of divisors, is also a power of two.

Therefore, if the sum of all positive divisors is a power of two, the number of divisors must also be a power of two. Thus, we conclude:

$$\boxed{\text{The number of divisors is a power of two.}}$$