How to find all rational solutions of $\ x^2 + 3y^2 = 7 $?

I knew that for $ x^2 + y^2 = 1$ the x and y can be expressed by introducing one more variable where $\ m=y/(x+1) $, then $\ x= 2m/(1+m^2) $ and $\ y= (1-m^2)/(1+m^2) $. What about $\ x^2 + 3y^2 = 7 $, should I divide the equation by 7 in order to get the 1 at the right hand side ? Then how to deal with the $\ 3y^2 $ ? Thank you!

The key here is to understand where the substitution $$m = \frac{y}{x + 1}$$ comes from. Geometrically, this variable represents the slope of the line between a point $(x, y)$ and the point $(-1, 0)$. What you are then doing is considering a rational slope $m$, taking the line $L$ through $(-1,0)$ of slope $m$, and solving for the second intersection point of this line with the conic $x^2+y^2=1$ (the first intersection point being $(-1,0)$).

This same approach works for any conic, as long as you have a single rational point on the conic to play the role of $(-1,0)$. In the case of $x^2+3y^2=7$, for instance, you could take $(2,1)$ as your initial point. So then you would define $$m = \frac{y - 1}{x - 2}$$ and solve for $(x,y)$ as the second point where the line $L$ through $(2,1)$ of slope $m$ intersects the conic $x^2+3y^2=7$.

Using the method of pg 7 of this paper on this related equation $$x^2+3y^2=7z^2 \quad \text{with initial solution} \quad (x,y,z)=(2,1,1)$$

A line $y=t(x-2)+1$, which will cut through the ellipse $x^2 + 3y^2 = 7$ at rational points if $t$ is rational.... when substituted into the ellipse yields:

$$\begin{align} x^2+3\left[t(x-2)+1\right]^2&=7 \\ x^2+3\left[t^2(x^2-4x+4)+2t(x-2)+1\right]&=7 \\ (1+3t^2)x^2+(-12t^2+6t)x+(12-12t+3-7)&=0 \\ \text{vieta: the two roots,} \quad x_1, x_2 \quad \text{are such that} \quad -(x_1+x_2)&=\frac{-12t^2+6t}{1+3t^2} \\ \text{since} \quad x_1=2, \quad \text{we have} \quad x_2(t)=\frac{12t^2-6t}{1+3t^2}-2 &=\frac{6t^2-6t-2}{1+3t^2} \\ \text{substituting this} \quad x(t) \quad \text {into the line:} \quad y&=t\left(\frac{6t^2-6t-2}{1+3t^2}-2\right)+1 \\ y(t)=t\left(\frac{-6t-4}{1+3t^2}\right) + \frac{1+3t^2}{1+3t^2}&=\frac{-3t^2-4t+1}{1+3t^2} \end{align}$$

Letting $y(t)\to |y(t)|$, Solution set with one parameter:

$$\begin{cases} x(t)&=\frac{6t^2-6t-2}{1+3t^2} \\ y(t)&=\frac{3t^2+4t-1}{1+3t^2} \end{cases}$$

$t$ was rational, so let $t=\frac{m}{n}$

Solution set now with two parameters: $$\begin{align} x(m,n)&=\frac{6m^2-6mn-2n^2}{n^2+3m^2} \\ y(m,n)&=\frac{3m^2+4mn-n^2}{n^2+3m^2} \end{align}$$

This is not (yet) an answer, but an extended comment, just to have some numerical solutions to crosscheck given analytic solutions in other answers and comments. I document only solutions here with $\gcd(x,y,z)=1$.
Here the third variable $z$ stems from rewriting $ x^2+3 y^2 = 7 $ where $ (x,y) \in \mathbb Q$ by $ (x/z)^2+3 (y/z)^2 = 7 $ and then $$ x^2+3 y^2 = 7z^2 \qquad (x,y) \in\mathbb Z, z \in \mathbb N^+ \tag 1 $$

Let $M= \begin{bmatrix} 8&0&3\\0&1&0\\21&0&8 \end{bmatrix}$ a transfer-matrix to recurse from one solution to another with constant $y$.
$\qquad \\ \\$

Beginning at $A_{1,0}=[2,1,1]=[x,y,z]$ we get the infinite set of solutions

$A_{1,0} \cdot M^i =A_{1,i} \qquad \qquad $ for i=-5..5

   i       x    y      z
 -------------------------------
 ...      ...   ...    ...
  -5   -331973  1    125474
  -4    -20830  1      7873
  -3     -1307  1       494
  -2       -82  1        31
  -1        -5  1         2
   0         2  1         1
   1        37  1        14
   2       590  1       223
   3      9403  1      3554
   4    149858  1     56641
   5   2388325  1    902702
  ...     ...   ...    ...

Sidenote: as you might observe, we have also another recursion inside the columns of $x_i$ and $z_i$ : $$x_{2+i}=16 x_{1+i} - 1 x_i$$

Using this matrix allows to shift from the set of solutions $A_{k,i}$ to $A_{3k,i}$ So $A_{1,0} \cdot Q = A_{3,0}$ which is numerically $[2,1,1] \cdot Q = [1,3,2]$ and

$A_{3,0} \cdot M^i =A_{3,i} \qquad \qquad $ for i=-5..5

   i       x    y      z
 -------------------------------
 ...      ...   ...    ...
  -5  -2206210  3   833869
  -4   -138431  3    52322
  -3     -8686  3     3283
  -2      -545  3      206
  -1       -34  3       13
   0         1  3        2
   1        50  3       19
   2       799  3      302
   3     12734  3     4813
   4    202945  3    76706
   5   3234386  3  1222483

General remark: because all $x,y,z$ occur as squares in the basic equation, we have always $\pm x$,$\pm y$ and $\pm z$ as valid solutions; however the matrix-formulae require meaningful attributing of signs.