I make the following conjecture from an answer.
conjecture
Suppose $N$ could be expressed as
$$ N= p^2\pm q,\ \ \ \ (p,q \in \mathbb{N},\ 1 \lt q \lt p,\ q\mid 2p) $$
In this case, the fundamental solution of $x^2-Ny^2=1$ is obtained by the following expression. (double sign in same order)
$$ x_0=\frac{2p^2}{q} \pm 1,\ \ y_0=\frac{2p}{q} $$
$(x_0,y_0)$ is one of the solutions of $x^2-Ny^2=1$ because it can be calculated as follows.
\begin{eqnarray*} x_0^2-Ny_0^2&=& \left(\frac{2p^2}{q} \pm 1\right)^2-(p^2\pm q)\cdot \left(\frac{2p}{q}\right)^2\\ &=& \frac{4p^4}{q^2} \pm \frac{4p^2}{q} +1 -\frac{4p^4}{q^2} \mp\frac{4p^2}{q}\\ &=& 1 \end{eqnarray*}
But, I don't know how to prove that it is the fundamental solution.
concrete example
Let $N$ be $75$.
The expression using the square number of the nearest neighbors is
$$ 75 = 9^2-6 $$
$p=9,\ q=6$ satisfy $1 \lt q \lt p$ and $\ q\mid 2p$.
Therefore,
\begin{eqnarray*} x_0&=&\frac{2p^2}{q} - 1,\ \ \ \ y_0&=&\frac{2p}{q} \\ &=& \frac{2\cdot 9^2}{6} -1 &=& \frac{2\cdot 9}{6} \\ &=& 26 &=& 3 \end{eqnarray*}
$(x_0, y_0) = (26, 3)$ is the fundamental solution of $x^2-75y^2=1$.