. To find solutions of $x^2 - 6x - 13 \equiv 0 \mod 127$, we must write $x^2-6x - 13$ as a square of a linear term, plus a constant. It's seen easily that $x^2 - 6x - 13 = (x-3)^2 - 22$. Hence, the congruence is equivalent to $(x-3)^2 \equiv 22 \mod 127$.
Now, we must check if $22$ is a quadratic residue mod $127$. For this, we can use the Legendre symbol, whose notation I will keep the same as the binomial, so do not get confused.
$$
\binom{22}{127} = \color{green}{\binom{11}{127}}\color{red}{\binom{2}{127}} = \color{green}{-\binom{127}{11}} \times \color{red}1
$$
where the terms with same color on the LHS and RHS are equal. The green equality comes by quadratic reciprocity and the second comes by the fact that $\binom{2}{p}$ is well known by the remainders which $p$ leaves when divided by $8$.
Now, we can do:
$$
-\binom{127}{11} = -\binom{6}{11} = -\color{green}{\binom{2}{11}}\color{red}{ \binom{3}{11}} = -\color{green}{(-1)}\color{red}{(1)} = 1
$$
Again, the colored terms on the LHS and RHS are equal because the quantities $\binom 2p$ and $\binom 3p$ are well known.
Since we have obtained that the Legendre symbol is $1$, this implies the existence of a solution, and therefore two.
The question, though, is how to compute them. I do not know of any method other than brute force, unfortunately. However, our reward for writing $(x-3)^2 \equiv 22 \mod 127$ is that we basically only need to look for squares of the form $127k + 22$ to find a solution, rather than having to substitute values of $x$ into the expression $x^2-6x-13$ each time.
A brute force : the series $127k + 22 $ goes like : $22,149,276,403,530,657,\color{blue}{784},...$
lo and behold, $784 = 28^2$, hence this gives $x = 31$. Now, note that the congruence actually has two solutions, one given by $127 - 28 = 99$. You can check that $9801 = 99^2 = 22 + 127 \times 77$.
Hence, we get two solutions of $x$, namely $x= 31,102$.
Easy way: $ $ applying my denesting formula we can compute $\color{#0a0}{\sqrt x}\,$ by completing a square
$\qquad \qquad\ 34x = x^2+1 \iff 36x = (x\!+\!1)^2\!\!\iff x = \left(\!\color{#0a0}{\dfrac{x\!+\!1}6}\!\right)^{\!2}$
Remark $ $ So $\:\!x\:\!$ has a $\:\!\rm\color{#0a0}{square\ root}\:\!$ in any ring where $\color{#0a0}{1/6}\,$ exists, e.g. $\:\!\Bbb Z_n\:\!$ for all $\:\!n\:\!$ coprime to $\:\!6,\,$ e.g. OP's $\,n=$ prime $p\equiv 7\pmod{\!8},\,$ where $\,p\neq 2,3\Rightarrow p\,$ coprime to $\:\!6$.
Best Answer
If $p$ is an odd prime and $-34\equiv a^2\pmod{p}$ for some $a$ then $$0^4-17\equiv2\left(\tfrac{a}{2}\right)^2\pmod{p}.$$
The linked question gives solutions to the congruence $$a^4-17b^4\equiv2c^2\pmod{p},$$ for every prime number $p$. Moreover, all solutions given there have $b\not\equiv1\pmod{p}$. So for $$x\equiv ab^{-1}\pmod{p} \qquad\text{ and }\qquad y\equiv cb^{-2}\pmod{p},$$ we have the following chain of congruences mod $p$: $$x^4-17\equiv a^4b^{-4}-17\equiv b^{-4}(a^4-17b^4)\equiv b^{-4}(2c^2)\equiv 2(cb^{-2})^2\equiv2y^2\pmod{p}.$$ This shows that every solution $(a,b,c)$ to the linked question with $b\not\equiv0\pmod{p}$ yields a solution $(x,y)=(ab^{-1},cb^{-2})$ to your question.