Simon would be pleased

We will prove that there exists no ordered pairs $(m, n)$ such that $8mn + 6m - n = p^2 + 1$ , for some positive integer $p$ .

Suppose this is possible i.e. there exists $(m, n)$ such that $8mn + 6m - n = p^2 + 1$ . By doing some algebraic manipulation, we get

$\begin{aligned} 8mn + 6m - n - 1 &= p^2 \\ 4(8mn + 6m - n - 1) &= (2p)^2 \\ 32mn + 24m - 4n - 4 &= (2p)^2 \\ 32mn + 24m - 4n - 3 &= (2p)^2 + 1 \\ 8m(4n + 3) - 1(4n + 3) &= (2p)^2 + 1 \\ (8m - 1)(4n + 3) &= (2p)^2 + 1. \end{aligned}$

Now, notice that the left hand side of this equation has at least one prime factor of the form $4k + 3$ . However, all prime factors that divide a number of the form $x^2 + 1$ has the form $4k + 1$ , a contradiction, which means that there are $\boxed{0}$ ordered pairs $(m, n)$ that satisfy the condition.