Note that your previous question had already answers at Which triangular numbers are also squares?
The part you may not know well enough is solving $x^2 - 8 y^2 = 1.$ Not sure how much to say...
jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
Automorphism matrix:
3 8
1 3
Automorphism backwards:
3 -8
-1 3
3^2 - 8 1^2 = 1
x^2 - 8 y^2 = 1
Sat Apr 1 10:00:02 PDT 2017
x: 3 y: 1 ratio: 3 SEED BACK ONE STEP 1 , 0
x: 17 y: 6 ratio: 2.83333
x: 99 y: 35 ratio: 2.82857
x: 577 y: 204 ratio: 2.82843
x: 3363 y: 1189 ratio: 2.82843
x: 19601 y: 6930 ratio: 2.82843
x: 114243 y: 40391 ratio: 2.82843
x: 665857 y: 235416 ratio: 2.82843
x: 3880899 y: 1372105 ratio: 2.82843
x: 22619537 y: 7997214 ratio: 2.82843
Sat Apr 1 10:01:02 PDT 2017
x^2 - 8 y^2 = 1
jagy@phobeusjunior:~$
You probably do not know matrices. Given one solution $x^2 - 8 y^2 = 1,$ the next solution is
$$ (x,y) \mapsto (3x+8y, x + 3y) $$
By the Cayley Hamilton Theorem, this tells us that, with $x_0 = 1, y_0 = 0,$
$$ x_{j+2} = 6 x_{j+1} - x_j, $$
$$ y_{j+2} = 6 y_{j+1} - y_j. $$
You can also confirm these formulas yourself, without using any matrices. Suggest you do that.
From your question yesterday:
I don't want the actual answer with the solution but I'd like to solve
the problem myself : I need some clues to take me on the right track.
Really brief tutorial: given $n > 0$ not a square, to find all integer solutions to $x^2 - n y^2 = 1$ with $x,y \geq 0,$ first find the smallest $u,v > 0$ such that $u^2 - n v^2 = 1.$ This is often called the fundamental solution. They give a table of fundamental solutions for $n \leq 128$ at TABLE
Given any solution $(x,y),$
the next solution is
$$ (x,y) \mapsto (ux+nvy, vx + uy) $$
By the Cayley Hamilton Theorem, this tells us that, with $x_0 = 1, y_0 = 0,$ then $x_1 = u, y_1 = v,$
$$ x_{j+2} = 2u x_{j+1} - x_j, $$
$$ y_{j+2} = 2u y_{j+1} - y_j. $$