$\newcommand{\Leg}[2]{\left(\frac{#1}{#2}\right)} $
Note: In this question $ a \mid b $ denotes a divides b and $\Leg a b$ denotes Legendre's symbol.
Theorem 9.12 in Introduction to Analytic Number Theory states:
The Diophantine equation $$ y^2 = x^3 + k \tag{10} $$ has no solutions if $ k $ has the form $$ k = (4n - 1)^3 - 4m^2, \tag{11} $$ where $ m $ and $ n $ are integers such that no prime $ p \equiv -1 \pmod{4} $ divides $m$.
The proof is presented below. I have understood most of the proof except one sentence (the sentence before the last sentence).
We assume a solution $x$, $y$ exists and obtain a contradiction by considering the equation modulo 4. Since $ k \equiv 1 \pmod{4}$ we have $$ y^2 \equiv x^3 - 1 \pmod{4}. \tag{12}$$ Now $y^2 \equiv 0$ or $1$ $\pmod{4}$ for every y, so (12) cannot be satisfied if $x$ is even or if $x \equiv -1 \pmod{4}.$ Therefore we must have $x \equiv 1 \pmod{4}$. Now let $$ a = 4n - 1 $$ so that $ k = a^3 - 4m^2 $, and write (10) in the form $$ y^2 + 4m^2 = x^3 + a^3 = (x + a)(x^2 - ax + a^2). \tag{13}$$ Since $ x \equiv 1 \pmod{4} $ and $ a \equiv -1 \pmod{4}$ we have $$ x^2 - ax + a^2 \equiv 1 - a + a^2 \equiv -1 \pmod{4}. \tag{14} $$ Hence $ x^2 - ax + a^2 $ is odd, and (14) shows that all its prime factors cannot be $ \equiv 1 \pmod{4}.$ Therefore some prime $p \equiv -1 \pmod{4}$ divides $ x^2 - ax + a^2,$ and (13) shows that this also divides $y^2 + 4m^2.$ In other words, $$ y^2 \equiv -4m^2 \pmod{p} \text{ for some } p \equiv -1 \pmod{4}. \tag{15} $$ But $ p \nmid m $ by hypothesis, so $\Leg{-4m^2}{p} = \Leg{-1} p = -1, $ contradicting (15). This proves that the Diophantine equation (10) has no solutions when $k$ has the form (11).
The part that I do not understand is this:
But $ p \nmid m $ by hypothesis, so $\Leg{-4m^2} p = \Leg{-1}p $
Why does $ p \nmid m $ imply $\Leg{-4m^2} p = \Leg{-1} p $?
As mentioned in the comments, this just follows from the multiplicative nature of the Legendre symbol. Indeed,
$$\left(\frac {-4m^2}p\right)=\left(\frac {-1}p\right) \times \left(\frac {4m^2}p\right)=\left(\frac {-1}p\right)$$
Since $4m^2$ is, clearly, a non-zero square (trusting that $p>2$).