For our purpose,let $a,b,c$ and $x\gt2$ be natural numbers such that the positive integers $a,b$ and $c$ form a special pythagorean triple $(a,b,c)$,then it is conjectured that the following is true
$$G(x)=\cfrac{2}{3x+\cfrac{(-1)(7)} {9x+\cfrac{(2)(10)}{15x+\cfrac{(5)(13)}{21x+\cfrac{(8)(16)}{27x+\ddots}}}}}$$
$$G(x)=\frac{a}{2b}+\frac{c}{b}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{c+a}{2c}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{c-a}{2c}\right) \right]$$
where,
$$a=4x(x^2-1)\\b = x^4-6x^2+1\\c=(x^2+1)^2$$
such that,
$$a^2+b^2=c^2=(x^2+1)^4$$
Corollaries
Here are some of its closed forms with their related not necessarily primitive pythagorean triples next to them
$$G(3)=\frac{12}{7}+\frac{25}{7}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{49}{50}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{1}{50}\right) \right] ; (7,24,25)$$
$$G(4)=\frac{120}{161}+\frac{289}{161}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{529}{578}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{49}{578}\right) \right] ;(161,240,289)$$
$$G(5)=\frac{60}{119}+\frac{169}{119}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{289}{338}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{49}{289}\right) \right] ;(119,120,169)$$
$$G(6)=\frac{420}{1081}+\frac{1369}{1081}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{2209}{2738}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{529}{2738}\right) \right] ;(840,1081,1369)$$
$$G(7)=\frac{168}{527}+\frac{625}{527}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{961}{1250}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{289}{1250}\right) \right] ;(336,527,625)$$
$$G(8)=\frac{1008}{3713}+\frac{4225}{3713}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{6241}{8450}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{2209}{8450}\right) \right] ;(2016,3713,4225)$$
$$G(9)=\frac{360}{1519}+\frac{1681}{1519}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{2401}{3362}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{961}{3362}\right) \right] ;(720,1519,1681)$$
$$G(10)=\frac{1980}{9401}+\frac{10201}{9401}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{14161}{20402}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{6241}{20402}\right) \right] ;(3960,9401,10201)$$
$$G(11)=\frac{660}{3479}+\frac{3721}{3479}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{5041}{7442}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{2401}{7442}\right) \right] ;(1320,3479,3721)$$
$$G(12)=\frac{3432}{19873}+\frac{21025}{19873}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{27889}{42050}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{14161}{42050}\right) \right] ;(6864,19873,21025)$$
and so on for all values of $x\gt2$...
Question :Can the conjecture be rigorously proved,so that the connection between the continued fraction and the primitive pythagorean triples can be established?
(Too long for a comment.)
From your other post, we have,
$$G_1(x,n)=\cfrac{1}{2x+\cfrac{(-1)(-1+n)} {6x+\cfrac{(1)(1+n)}{10x+\cfrac{(3)(3+n)}{14x+\cfrac{(5)(5+n)}{18x+\ddots}}}}}\tag1$$
with $2v+1 =-1,1,3,5,\dots$ The special case $n=6$,
$$G_1(x,\color{brown}6)=\frac{1}{3b}\left(a+\sqrt{c^3}\right)\tag2$$
where,
$$a=-x^3+3x,\;b=3x^2-1,\;c=x^2+1$$
So $G_1(x,\color{brown}6)$ is a root of a quadratic. Note that,
$$\color{blue}{a^2+b^2=c^3=(x^2+1)^3}\tag3$$
From this post, we have the analogous,
$$G_2(x,n)=\cfrac{1}{3x+\cfrac{(-1)(-1+n)} {9x+\cfrac{(2)(2+n)}{15x+\cfrac{(5)(5+n)}{21x+\cfrac{(8)(8+n)}{27x+\ddots}}}}}\tag4$$
with $3v+2 =-1,2,5,8,\dots$ The special case $n=8$,
$$G_2(x,\color{brown}8)=\frac{a}{4b}+\frac{c}{2b}\left[ {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{c+a}{2c}\right) - {}_2F_1\left( -\frac13, \frac13 ; \frac12; \frac{c-a}{2c}\right) \right]\tag{5a}$$
Equivalently,
$$G_2(x,\color{brown}8)=\frac{1}{4b}\left(a+c\Big(u+\frac{1}{u}\Big)\right)\tag{5b}$$
where $\displaystyle u =e^{4\pi\, i /3}\left(\frac{a+b\,i}{c}\right)^{1/3}$ and,
$$a = 2m,\;b=m^2-1,\;c=m^2+1,\quad\text{with}\quad m=\frac{x^2-1}{2x}$$
So $G_2(x,\color{brown}8)$ is a root of a cubic. Note that,
$$\color{blue}{a^2+b^2=c^2=\left(\frac{x^2+1}{2x}\right)^4}\tag6$$
Another cfrac with a closed-form, presumably a root of a quartic?