Let $s = m + n$ and $p = mn$. Then the equation becomes:
$$s^2 + p + 1 = N s$$
So,
$$p = N s - s^2 - 1$$
Recall that $m$ and $n$ are roots of the quadratic equation $x^2 - s x + p = 0$.

The discriminant is:
$$D = s^2 - 4p = s^2 - 4(N s - s^2 - 1) = s^2 - 4N s + 4s^2 + 4$$
$$= 5s^2 - 4N s + 4$$
For $m$ and $n$ to be integers, $D$ must be a perfect square.

Let $D = t^2$ for some integer $t$.
So,
$$5s^2 - 4N s + 4 = t^2$$
This is a quadratic in $s$ for each integer $t$.

For each integer $t$, the equation
$$5s^2 - 4N s + 4 - t^2 = 0$$
has discriminant in $s$:
$$\Delta = (4N)^2 - 4 \cdot 5 \cdot (4 - t^2) = 16N^2 - 20(4 - t^2) = 16N^2 - 80 + 20 t^2$$
For sufficiently large $|t|$, this is positive, so there are integer solutions for $s$ for infinitely many $t$.

Alternatively, fix $s$ to be any integer, then $p = N s - s^2 - 1$ is also integer, and $m$ and $n$ are roots of $x^2 - s x + p = 0$.

The roots are
$$m, n = \frac{s \pm \sqrt{s^2 - 4p}}{2}$$
So $s^2 - 4p$ must be a perfect square.
But $s^2 - 4p = s^2 - 4(N s - s^2 - 1) = 5s^2 - 4N s + 4$

So for each $s$ such that $5s^2 - 4N s + 4$ is a perfect square, we get integer solutions.

But for each $k$, let $s$ be such that $5s^2 - 4N s + 4 = k^2$ for some integer $k$.
This is a quadratic in $s$:
$$5s^2 - 4N s + 4 - k^2 = 0$$
For each integer $k$, this quadratic has integer solutions for $s$ for infinitely many $k$.

Therefore, there are infinitely many integer solutions $(m, n)$ to the given equation for any integer $N$.