To find all pairs of positive integers $(m, n)$ such that $mn - 1$ divides $(n^2 - n + 1)^2$, we need to analyze the given condition and derive the solutions.

First, let's denote $d = mn - 1$. We need $d$ to divide $(n^2 - n + 1)^2$. This implies:
$$d \mid (n^2 - n + 1)^2.$$

We start by considering the trivial solution $(m, n) = (2, 2)$:
$$2 \cdot 2 - 1 = 3 \quad \text{and} \quad (2^2 - 2 + 1)^2 = 3^2 = 9,$$
which satisfies $3 \mid 9$.

Next, we explore other potential solutions. Suppose $(m, n)$ is a solution with $2 < n < m$. We need to find $k < n$ such that $(n, k)$ is also a solution. This requires:
$$(n^2 - n + 1)^2 \equiv -1 \pmod{n}.$$

We analyze the inequalities:
$$n^2 - 1 \leq \frac{(n^2 - n + 1)^2}{mn - 1} \leq \frac{(n^2 - n + 1)^2}{n^2 - 1}.$$

Simplifying, we get:
$$\frac{(n^2 - n + 1)^2}{n^2 - 1} \geq n^2 - 1 \implies n^2 - n + 1 \geq n^2 - 1 \implies n \leq 2.$$

This implies that the solutions are limited to specific cases. By fixing initial values $a_0$ and $b_0$ with $b_0 \leq 2$, we can generate all solutions using the recurrence relation:
$$(a_i, b_i)_{i=0}^{\infty} \quad \text{where} \quad a_i = b_{i+1}, \quad b_i = \frac{1}{b_{i+1}} \left[ \frac{(b_{i+1}^2 - b_{i+1} + 1)^2}{a_{i+1} b_{i+1} - 1} + 1 \right] \quad \forall \ i \in \mathbb{N}^*.$$

By working through this recurrence, we find that the solutions are:
$$(m, n) = (2, 2) \quad \text{and} \quad (m, n) = (b_{\ell-1}, b_{\ell}), \ (b_{\ell}, b_{\ell+1}) \quad \forall \ \ell \in \mathbb{N}^*.$$

Cleaning up the recurrence relation:
$$b_{i+2} = \frac{b_{i+1}^3 - 2b_{i+1}^2 + 3b_{i+1} + b_i - 2}{b_i b_{i+1} - 1} \quad \forall \ i \in \mathbb{N}^*, \quad b_0 = 2, \quad b_1 = 5,$$
we get the sequence:
$$(b_i)_{i=0}^{\infty} = 2, 5, 10, 17, \ldots,$$
which can be generalized as:
$$b_i = (i + 1)^2 + 1.$$

Thus, the pairs $(m, n)$ that satisfy the condition are:
$$(m, n) = (2, 2) \quad \text{and} \quad (m, n) = ((i+1)^2 + 1, (i+2)^2 + 1) \quad \forall \ i \in \mathbb{N}.$$

The answer is: $\boxed{(2,}$ 2) $\text{}$ and } ((i+1)^2 + 1, (i+2)^2 + 1) $\text{}$ for all } i ${\in \mathbb{N}}.$