If $x^2-991y^2=1$, then we have that $ \begin{align} \left(\frac{1}{y}\right)^2 &=\left(\frac{x}{y}\right)^2-991\\ &=\left(\frac{x}{y}-\sqrt{991}\right)\left(\frac{x}{y}+\sqrt{991}\right)\tag{1} \end{align} $ Since $\frac{x}{y}\stackrel{.}{=}\sqrt{991}\stackrel{.}{=}31.5$, we get that $ \left|\frac{x}{y}-\sqrt{991}\right|\stackrel{.}{=}\frac1{63}\frac1{y^2}\tag{2} $ As described in Section $5$ of this paper, to get a rational approximation as close as $(2)$, $\dfrac{x}{y}$ must be a convergent of the continued fraction for $\sqrt{991}$. Since the next continuant, c_n, satisfies $ \frac1{c_n+2}\frac1{y^2}\lt\left|\frac{x}{y}-\sqrt{991}\right|\le\frac1{c_n}\frac1{y^2}\tag{3} $ the next continuant should be within about $1$ of $62$.
The continued fraction for an algebraic number of degree $2$ will eventually repeat. The continued fraction for $\sqrt{991}$ is $ (31, [2, 12, 10, 2, 2, 2, 1, 1, 2, 6, 1, 1, 1, 1, 3, 1, 8, 4, 1, 2, 1, 2, 3, 1, 4, 1, 20, 6, 4, 31, 4, 6, 20, 1, 4, 1, 3, 2, 1, 2, 1, 4, 8, 1, 3, 1, 1, 1, 1, 6, 2, 1, 1, 2, 2, 2, 10, 12, 2, 62]) $ where the sequence in brackets repeats.
The only continuant within $1$ of $62$ is the $62$. Thus, the first positive solution corresponds to the rational approximation to $\sqrt{991}$ with the continued fraction $ (31, 2, 12, 10, 2, 2, 2, 1, 1, 2, 6, 1, 1, 1, 1, 3, 1, 8, 4, 1, 2, 1, 2, 3, 1, 4, 1, 20, 6, 4, 31, 4, 6, 20, 1, 4, 1, 3, 2, 1, 2, 1, 4, 8, 1, 3, 1, 1, 1, 1, 6, 2, 1, 1, 2, 2, 2, 10, 12, 2) $ which is $ \color{#C00000}{\frac{x_1}{y_1}}=\color{#C00000}{\frac{379516400906811930638014896080}{12055735790331359447442538767}}\tag{4} $ The next solution comes from the next time the next continuant is $62$; that is, for the rational approximation to $\sqrt{991}$ with the continued fraction $ (31, 2, 12, 10, 2, 2, 2, 1, 1, 2, 6, 1, 1, 1, 1, 3, 1, 8, 4, 1, 2, 1, 2, 3, 1, 4, 1, 20, 6, 4, 31, 4, 6, 20, 1, 4, 1, 3, 2, 1, 2, 1, 4, 8, 1, 3, 1, 1, 1, 1, 6, 2, 1, 1, 2, 2, 2, 10, 12, 2, 62, 2, 12, 10, 2, 2, 2, 1, 1, 2, 6, 1, 1, 1, 1, 3, 1, 8, 4, 1, 2, 1, 2, 3, 1, 4, 1, 20, 6, 4, 31, 4, 6, 20, 1, 4, 1, 3, 2, 1, 2, 1, 4, 8, 1, 3, 1, 1, 1, 1, 6, 2, 1, 1, 2, 2, 2, 10, 12, 2) $ which is $ \color{#C00000}{\frac{x_2}{y_2}}=\color{#C00000}{\frac{288065397114519999215772221121510725946342952839946398732799}{9150698914859994783783151874415159820056535806397752666720}}\tag{5} $ We can get all the solutions, starting with $(x_0,y_0)=(1,0)$ and $(4)$, using $ \begin{align} x_n&=759032801813623861276029792160\,x_{n-1}-x_{n-2}\\ y_n&=759032801813623861276029792160\,y_{n-1}-y_{n-2} \end{align}\tag{6} $