To find all ordered pairs of positive integers $(m, n)$ such that $mn-1$ divides $m^2 + n^2$, we start by considering the condition:
$$\frac{m^2 + n^2}{mn - 1} = c \quad \text{where} \quad c \in \mathbb{Z}.$$
This implies:
$$m^2 + n^2 = c(mn - 1).$$
Rewriting, we get:
$$m^2 - cmn + n^2 + c = 0.$$

Let $(m, n)$ be a solution where $m + n$ is minimized. If $(m, n)$ is a solution, then $(m', n)$ must also be a solution, where:
$$m' = cn - m = \frac{n^2 + c}{m}.$$
Since $m'$ is positive, $cn - m > 0$, and since $m'$ is an integer, $cn - m \ge 1$.

Assuming $m \ne n$ and without loss of generality, $m > n$, we claim $n = 1$. For contradiction, assume $n \ne 1$. Then $n > 1$ implies $m > n > 1$. By minimality of $m + n$, we must have:
$$m + n \le m' + n \implies m \le m'.$$
However, since $m > n > 1$:
$$n(m - n) \ge 2 \implies mn - 2 \ge n^2 \implies m(mn - 2) \ge mn^2 > n^3,$$
$$n(m^2 - n^2) > 2m \implies m^2n > 2m + n^3,$$
$$2m^2n - 2m > m^2n + n^3 \implies 2m(mn - 1) > n(m^2 + n^2) \implies m > cn - m = m',$$
a contradiction. Thus, $n = 1$.

For $n = 1$, we have:
$$\frac{m^2 + 1}{m - 1} = (m + 1) + \frac{2}{m - 1}.$$
This is an integer if and only if $m - 1$ divides 2. Hence, $m = 2$ or $m = 3$. Therefore, the solutions for $m > n$ are $(2, 1)$ and $(3, 1)$.

Since the expression is symmetric in $m$ and $n$, the pairs $(m, n)$ that satisfy $mn - 1 \mid m^2 + n^2$ are:
$$\boxed{(2, 1), (3, 1), (1, 2), (1, 3)}.$$