Find all ordered pairs $ (m,n)$ where $ m$ and $ n$ are positive integers such that $ \frac {n^3 \plus{} 1}{mn \minus{} 1}$ is an integer.
1. We start with the given condition that $\frac{n^3 + 1}{mn - 1}$ is an integer. This implies that $mn - 1$ divides $n^3 + 1$.
   
2. Note that $\gcd(mn-1, n) = 1$ because $mn-1$ and $n$ are coprime. This is because $mn-1$ is not divisible by $n$.

3. We can rewrite the expression as:
   \[
   \frac{n^3 + 1}{mn - 1} = \left(\frac{n^3 + 1 + mn - 1}{mn - 1}\right) - 1 = \frac{n^3 + mn}{mn - 1} - 1
   \]
   Simplifying further, we get:
   \[
   \frac{n^3 + mn}{mn - 1} = \frac{n(n^2 + m)}{mn - 1}
   \]
   Since $\frac{n(n^2 + m)}{mn - 1}$ is an integer, it implies that $mn - 1$ divides $n^2 + m$.

4. Now, consider the expression $\frac{n^2 + m}{mn - 1}$. We can rewrite it as:
   \[
   \frac{n^2 + m}{mn - 1} = \frac{n(mn - 1) + m^2 + n}{mn - 1} = n + \frac{m^2 + n}{mn - 1}
   \]
   Since $n + \frac{m^2 + n}{mn - 1}$ is an integer, it implies that $mn - 1$ divides $m^2 + n$.

5. We now have two conditions:
   \[
   mn - 1 \mid n^2 + m \quad \text{and} \quad mn - 1 \mid m^2 + n
   \]

6. We can derive bounds for $m$ and $n$:
   \[
   \frac{n^2 + 1}{n - 1} \geq m \quad \text{and} \quad \frac{m^2 + 1}{m - 1} \geq n
   \]
   Simplifying these inequalities:
   \[
   m \leq \frac{n^2 + 1}{n - 1} = n + 1 + \frac{2}{n - 1} < n + 2 \quad \text{for } n > 3
   \]
   \[
   n \leq \frac{m^2 + 1}{m - 1} = m + 1 + \frac{2}{m - 1} < m + 2 \quad \text{for } m > 3
   \]

7. Hence, we have:
   \[
   n < m + 2 < n + 4 \implies m = n - 1, n, n + 1
   \]
   We need to manually check values when $m, n \leq 3$.

8. Checking manually for $m, n \leq 3$:
   - $(m, n) = (1, 2)$: $\frac{2^3 + 1}{1 \cdot 2 - 1} = \frac{9}{1} = 9$ (integer)
   - $(m, n) = (1, 3)$: $\frac{3^3 + 1}{1 \cdot 3 - 1} = \frac{28}{2} = 14$ (integer)
   - $(m, n) = (2, 1)$: $\frac{1^3 + 1}{2 \cdot 1 - 1} = \frac{2}{1} = 2$ (integer)
   - $(m, n) = (2, 2)$: $\frac{2^3 + 1}{2 \cdot 2 - 1} = \frac{9}{3} = 3$ (integer)
   - $(m, n) = (2, 5)$: $\frac{5^3 + 1}{2 \cdot 5 - 1} = \frac{126}{9} = 14$ (integer)
   - $(m, n) = (3, 1)$: $\frac{1^3 + 1}{3 \cdot 1 - 1} = \frac{2}{2} = 1$ (integer)
   - $(m, n) = (3, 5)$: $\frac{5^3 + 1}{3 \cdot 5 - 1} = \frac{126}{14} = 9$ (integer)
   - $(m, n) = (5, 2)$: $\frac{2^3 + 1}{5 \cdot 2 - 1} = \frac{9}{9} = 1$ (integer)
   - $(m, n) = (5, 3)$: $\frac{3^3 + 1}{5 \cdot 3 - 1} = \frac{28}{14} = 2$ (integer)

Thus, the ordered pairs $(m, n)$ that satisfy the condition are:
\[
\boxed{(1, 2), (1, 3), (2, 1), (2, 2), (2, 5), (3, 1), (3, 5), (5, 2), (5, 3)}
\]