Trying to solve a problem I meet on the way a question, easy maybe, that I can't solve for now. The Pell-Fermat equation $$x^2-dy^2=1 \hspace{2cm}(*)$$ where the integer $d$ is square-free, has in general infinitely many integer solutions for which the fundamental unit of $\mathbb Q(\sqrt d)$ is the key. In fact, all the integer solutions $(x_n,y_n)$ of $(*)$ are given by$$x_n+y_n\sqrt d=(x_1+y_1\sqrt d)^n;\space n\ge 1$$ where $ x_1+y_1\sqrt d$ is the mentioned fundamental unit.
The punctual question is, are there non-integer rational solutions $(x,y)$ with $xy\ne0$ for $x^2-3y^2=1?$
(All the integer solutions are given by $x_n+y_n\sqrt3=(2+\sqrt3)^n$)
This sort of thing is parametrized by a finite number of recipes of the Pythagorean triple type. It is not necessarily parametrized by just one recipe. A reference is Mordell, Diophantine Equations, page 47, Theorem 4.
Multiplying through be common denominator $w$ leads to integers $u,v,w$ with $$ u^2 - 3 v^2 = w^2 $$ or $$ u^2 = 3 v^2 + w^2 \; . \;$$
If, in turn, we demand $u,v,w$ coprime, ... with coprime integers $m,n:$
When $u$ is even we get all solutions by $$ u = 2 m^2 + 2mn + 2 n^2 \;, \; \; v = m^2 + 2mn \; , \; \; w = -m^2 + 2mn + 2 n^2 \; . \; $$
When $u$ is odd we get all solutions by $$ u = 3 m^2 + n^2 \;, \; \; v = 2mn \; , \; \; w = 3m^2 - n^2 \; . \; $$
In the first one we need $m$ odd, not sure about conditions $\pmod 3$
In the second one we need $m+n$ odd and $ n \neq 0\pmod 3$