$$ x=1+\cfrac{1}{1+\cfrac{1}{1+...}}\implies x=1+\frac{1}{x}\implies x=\frac{1\pm \sqrt{5}}{2} $$ Can the negative solution be considered as a solution? If yes, how is it possible to have a negative solution for a positive continued fraction? If no, how do we prove that it can't be a solution?
Edit 1: I want to understand the assumption we are considering while forming the equation which results in the "extraneous solution".
On
Look at what you've written (assuming the continued fraction converges): $$ x=1+\cfrac1{\color{#C00000}{1+\cfrac1{1+\cfrac1{1+\dots}}}}\tag{1} $$ implies that $$ x=1+\frac1{\color{#C00000}{x}}\tag{2} $$ which, in turn, implies that $$ x=\frac{1+\sqrt5}2\quad\text{or}\quad x=\frac{1-\sqrt5}2\tag{3} $$ By transitivity, $(1)\implies(3)$. Since $x$, as defined in $(1)$ is greater than $0$, we know that $x=\frac{1-\sqrt5}2$ is false. However, since $(3)$ is true, we must have $$ x=\frac{1+\sqrt5}2\tag{4} $$
No, the negative number is not a solution. You showed that if $x$ is equal to that fraction, then it is either $\frac{1+\sqrt 5}{2}$ or $\frac{1-\sqrt5}{2}$. You calculated possible candidates for solutions, not the solution itself.
You can prove that $x$ must be positive by simply arguing that $x$ is a limit of a sequence with only positive elements, so the limit (if it exists, which should also be proven) must be positive.