I'm the OP. I figured out the solution. The first supplement of quadratic reciprocity, we will denote (1S), and the second supplement we will denote (2S). Furthermore, we denote by $\textrm{sgn}(x)$ the sign of $x$ (i.e. $\pm 1$). In what follows, all $\pm $ signs in the same equation take on the same sign, and to denote where they have opposite sign, we use the $\mp$ sign. Lastly, I found a better version of (EQR) that implies not only (2S) and (QR), but also (1S). This version is as follows:
(EQR*) For any $a$ satisfying $p\equiv \pm q \bmod 4a$, we have $\left(\frac ap \right)=\textrm{sgn}(a)^{\frac{p-q}{2}}\left(\frac aq \right)$.
Now we will attempt to prove its equivalence to $(1S)\wedge (2S) \wedge (QR)$.
Claim. $(1S)\wedge(2S)\wedge (QR) \iff (EQR^*)$
Proof: ($\Longrightarrow$) Let $p\equiv \pm q \bmod 4a$ for some $a$. It suffices to prove for primes and $-1$. Note when $a>0$, we have $\textrm{sgn}(a)=+1$, so $\textrm{sgn}(a)^{\frac{p-q}{2}}=+1$. We set $p=\pm q+4ab$ for some $ b$.
If $a=-1$, then since $p-q$ is even, by $(1S)$ we have
\begin{align*}
\left( \frac ap \right) &= \left( \frac {-1}{p}\right)\\
&= (-1)^{\frac{p-1}{2}}\\
&=(-1)^{\frac{p-q}{2}}(-1)^{\frac{q-1}{2}}\\
&=\textrm{sgn}(a)^{\frac{p-q}{2}}\left( \frac aq \right)
\end{align*}
Now if $a=2$, then by $(2S)$ we have
\begin{align*}
\left( \frac ap \right) &=\left( \frac 2p \right)\\
&=(-1)^{\frac{p^2-1}{8}}\\
&=(-1)^{\frac{(\pm q +8b)^2-1}{8}}\\
&=(-1)^{\frac{q^2-1}{8}+\frac{\pm 16bq+16a^2b^2}{8}}\\
&=(-1)^{\frac{q^2-1}{8}}\\
&=\left( \frac 2q \right)\\
&=\left( \frac aq \right)
\end{align*}
Lastly, if $a$ is an odd prime, then by (QR) we have
\begin{align*}
\left( \frac ap \right) &= (-1)^{\frac{(p-1)(a-1)}{4}}\left( \frac pa \right) \\
&= (-1)^{\frac{(p-1)(a-1)}{4}}\left( \frac{\pm q +4ab}{a}\right) \\
&=(-1)^{\frac{(p-1)(a-1)}{4}}\left( \frac{\pm q}{a}\right) \\
&=(-1)^{\frac{(p-1)(a-1)}{4}}(\pm 1)^{\frac{a-1}{2}}\left( \frac qa \right) \\
&=(-1)^{\frac{(p-1)(a-1)}{4}}(-1)^{\mp \frac{(q-1)(a-1)}{4}}(\pm 1)^{\frac{a-1}{2}}\left( \frac aq \right)\\
&=(-1)^{\frac{pa-p-a+1\mp qa\pm q\pm a\mp 1}{4}}(\pm 1)^{\frac{a-1}{2}}\left( \frac aq \right) \\
&=(-1)^{\frac{(p\mp q)(a-1)-(a\mp a)+1\mp 1}{4}}(\pm 1 )^{\frac{2a-2}{4}}\left( \frac aq \right)\\
&=(-1)^{\frac{(p\mp q)(a-1)}{4}}(-1)^{\frac{-(a\mp a)+1\mp 1}{4}}(\pm 1 )^{\frac{2a-2}{4}}\left( \frac aq \right)\\
&=\underbrace{(-1)^{b(a-1)}}_{=+1}\underbrace{(-1)^{\frac{-(a\mp a)+1\mp 1}{4}}(\pm 1 )^{\frac{2a-2}{4}}}_{=+1}\left( \frac aq \right)\\
&=\left( \frac aq \right)
\end{align*}
Since $\left( \frac xp \right)$ and $\textrm{sgn}(x)$ are completely multiplicative functions, combining these results we know it holds for all $a$.
($\Longleftarrow$) Now let (EQR*) hold, and suppose $p=\pm q+4A$.
Now if $p\equiv 1 \bmod 4$, then $4\big| p-5$, and thus, $\left( \frac{-1}{p}\right)=\left( \frac {-1}{5}\right)=+1$. Furthermore, if $p\equiv 3 \bmod 4$, then $4 \big| p-3$, so $\left( \frac{-1}{p}\right)=\left( \frac{-1}{3}\right)=-1$. Therefore, $\left( \frac {-1}{p}\right)=(-1)^{\frac{p-1}{2}}$, so $(1S)$ holds.
Furthermore, since $p$ is odd, we have $p\equiv 1, 3, 5,$ or $ 7 \bmod 8$. Thus, $8\big| p-17, p-3, p-5, $ or $ p-7$. If $p\equiv \pm 1 \bmod 8$, then let $q =12\pm 5$. Then $\left( \frac 2p \right) = \left( \frac 2q\right)=+1$. Furthermore, if $p\equiv \pm 3 \bmod 8$, then let $q=4\mp 1$. Then $\left( \frac 2p \right) =\left( \frac 2q \right)=-1$. Therefore, $\left( \frac 2p \right) =(-1)^{\frac{p^2-1}{8}}$, so $(2S)$ holds.
Lastly, if $p=q+4A$, then
\begin{align*}
\left( \frac pq \right)&=\left( \frac{q+4A}{q}\right) \\
&=\left( \frac Aq \right) \\
&= \textrm{sgn}(A)^{\frac{p-q}{2}}\left( \frac Ap \right) \\
&=\textrm{sgn}(A)^{\frac{p-q}{2}}\left( \frac {-p+4A}{p}\right)\\
&=\textrm{sgn}(A)^{\frac{p-q}{2}}\left( \frac {-q}{p}\right) \\
&=\textrm{sgn}(A)^{\frac{p-q}{2}}(-1)^{\frac{p-1}{2}}\left( \frac qp \right) (*)
\end{align*}
If $A<0$, then we get $\textrm{sgn}(A)=-1$, so $\textrm{sgn}(A)^{\frac{p-q}{2}}(-1)^{\frac{p-1}{2}}=(-1)^{\frac{q-1}{2}}$. Otherwise, $\textrm{sgn}(A)^{\frac{p-q}{2}}(-1)^{\frac{p-1}{2}}=(-1)^{\frac{p-1}{2}}$. Since $p\equiv q \bmod 4$, know know $\frac{p-1}{2}\equiv\frac{q-1}{2}\equiv \frac{p-1}{2}\frac{q-1}{2}\bmod 2$, and therefore, ($*$) becomes $$\left( \frac pq \right) = (-1)^{\frac{(p-1)(q-1)}{4}}\left( \frac qp \right)$$
Similarly, if $p=-q+4A$, we know $A>0$. Also, we have either $p\equiv 1 \bmod 4$ or $q\equiv 1 \bmod 4$, so $\frac{(p-1)(q-1)}{4}$ is even. Then
\begin{align*}
\left( \frac pq \right) &=\left( \frac{-q+4A}{q}\right) \\
&=\left( \frac Aq \right) \\
&=\left( \frac Ap \right) \\
&=\left( \frac {-p+4A}{p}\right) \\
&=\left( \frac qp \right) \\
&=(-1)^{\frac{(p-1)(q-1)}{4}}\left( \frac qp \right)
\end{align*}
so $(QR)$ holds.
Best Answer
Your question is not difficult to answer, although some of the underlying proofs are technical (but have been known for over 100 years). In what sense does your friend "specialize" in higher reciprocity laws?
You ask if the $m$th power residue symbol "can be extended" to prime ideals, but you have things backwards: the $m$th power residue symbol is first defined on prime ideals. Letting $$ K = \mathbf Q(\zeta_m), $$ each prime ideal $\mathfrak p$ in $\mathbf Z[\zeta_m]$ that does not divide $m$ is unramified in $K$ and $x^m - 1$ is separable in $(\mathbf Z[\zeta_m]/\mathfrak p)[x]$. For each $\alpha \in \mathbf Z[\zeta_m]$ such that $\alpha \not\equiv 0 \bmod \mathfrak p$, $\alpha^{({\rm N}(\mathfrak p)-1)/m} \bmod \mathfrak p$ is an $m$th root of unity in the residue field $\mathbf Z[\zeta_m]/\mathfrak p$, so there is a unique $m$th root of unity $\zeta$ in $K$ such that $$ \alpha^{({\rm N}(\mathfrak p)-1)/m} \equiv \zeta \bmod \mathfrak p. $$ This $\zeta$ is defined to be the power residue symbol $(\frac{\alpha}{\mathfrak p})_m$. When $m = 2$, so $K = \mathbf Q$, this is the Legendre symbol $(\frac{a}{p})$ when $\mathfrak p = p\mathbf Z$ for an odd prime $p$ and integer $a$ not divisible by $p$.
The power residue symbol $(\frac{\alpha}{\mathfrak p})_m$ is a Frobenius element in a Kummer extension of $K$. Set $$ L = K(\sqrt[m]{\alpha}). $$ Since $K$ contains all the $m$th roots of unity, $L/K$ is abelian with a cyclic Galois group of order dividing $m$: ${\rm Gal}(L/K)$ emebds into $\mu_m$ by $\sigma \mapsto \sigma(\sqrt[m]{\alpha})/\sqrt[m]{\alpha}$. (That ratio is independent of the choice of $m$th root of $\alpha$, since different choices have a ratio in the $m$th roots of unity, which all lie in $K$ and thus are all fixed by $\sigma$.) Since the Galois group is abelian, primes in $K$ have well-defined Frobenius elements in $L$ when they are unramified in $L$.
Claim: When $\mathfrak p \nmid (\alpha)m$ in $\mathbf Z[\zeta_m]$, the Frobenius element ${\rm Frob}_{K(\sqrt[m]{\alpha})/K}(\mathfrak p)$ is identified with $(\frac{\alpha}{\mathfrak p})_m$ under the above embedding of ${\rm Gal}(K(\sqrt[m]{\alpha})/K)$ into $\mu_m$.
Proof: When $\mathfrak p \nmid (\alpha)m$, $\mathfrak p$ is unramified in $L$. Letting $\mathfrak P$ be a prime lying over $\mathfrak p$ in $\mathcal O_L$, by definition $$ {\rm Frob}_{L/K}(\mathfrak p)(\beta) \equiv \beta^{{\rm N}(\mathfrak p)} \bmod \mathfrak P $$ for all $\beta$ in $\mathcal O_L$. Taking $\beta = \sqrt[m]{\alpha}$, we get $$ {\rm Frob}_{L/K}(\mathfrak p)(\sqrt[m]{\alpha}) \equiv \sqrt[m]{\alpha}^{{\rm N}(\mathfrak p)} \bmod \mathfrak P. $$ Since $$ \sqrt[m]{\alpha}^{{\rm N}(\mathfrak p)} = \sqrt[m]{\alpha}^{({\rm N}(\mathfrak p)-1)}\sqrt[m]{\alpha} = (\sqrt[m]{\alpha}^m)^{({\rm N}(\mathfrak p)-1)/m}\sqrt[m]{\alpha} = \alpha^{({\rm N}(\mathfrak p)-1)/m}\sqrt[m]{\alpha}, $$ we have $$ {\rm Frob}_{L/K}(\mathfrak p)(\sqrt[m]{\alpha}) \equiv \alpha^{({\rm N}(\mathfrak p)-1)/m}\sqrt[m]{\alpha} \bmod \mathfrak P. $$ By the definition of the power-residue symbol, $\alpha^{({\rm N}(\mathfrak p)-1)/m} \equiv (\frac{\alpha}{\mathfrak p})_m \bmod \mathfrak p$. A congruence modulo $\mathfrak p$ is also true modulo $\mathfrak P$, so $$ {\rm Frob}_{L/K}(\mathfrak p)(\sqrt[m]{\alpha}) \equiv \left(\frac{\alpha}{\mathfrak p}\right)_m\sqrt[m]{\alpha} \bmod \mathfrak P. $$
Under the embedding of ${\rm Gal}(L/K)$ into $\mu_m$, the Frobenius element $\sigma := {\rm Frob}_{L/K}(\mathfrak p)$ is mapped to the ratio $\sigma(\sqrt[m]{\alpha})/\sqrt[m]{\alpha}$. Call this $m$th root of unity $\zeta$, so $\sigma(\sqrt[m]{\alpha}) = \zeta\sqrt[m]{\alpha}$. Therefore $$ \zeta\sqrt[m]{\alpha} \equiv \left(\frac{\alpha}{\mathfrak p}\right)_m\sqrt[m]{\alpha} \bmod \mathfrak P. $$ Since $\mathfrak p \nmid (\alpha)$, also $\mathfrak P \nmid (\sqrt[m]{\alpha})$ in $\mathcal O_L$, so we can divide both sides of the congruence by $\sqrt[m]{\alpha}$: $$ \zeta \equiv \left(\frac{\alpha}{\mathfrak p}\right)_m \bmod \mathfrak P. $$ Since $\mathcal O_L/\mathfrak P$ is a field of characteristic not dividing $m$, $m$th roots of unity that are congruent modulo $\mathfrak P$ must be equal, so $$ \zeta = \left(\frac{\alpha}{\mathfrak p}\right)_m. $$ In other words, for each prime ideal $\mathfrak p$ in $\mathbf Z[\zeta_m]$ not dividing $(\alpha)m$, the standard embedding of ${\rm Gal}(K(\sqrt[m]{\alpha})/K)$ into $\mu_m$ maps the Frobenius element of $\mathfrak p$ to the power residue symbol $(\frac{\alpha}{\mathfrak p})_m$. That finishes the proof of the claim.
The Chebotarev density theorem, in the case of an abelian Galois extension of degree $n$, says each element of the Galois group is a Frobenius element for a set of prime ideals in the base field with (natural) density $1/n$. In the case of the abelian extension $K(\sqrt[m]{\alpha})/K$, where $K = \mathbf Q(\zeta_m)$ and we embed the Galois group into $\mu_m$ in order to identity (all but finitely many) Frobenius elements with power residue symbols $(\frac{\alpha}{\mathfrak p})_m$, we can say that for each $m$th root of unity $\zeta$ that is in the image of the embedding of ${\rm Gal}(K(\sqrt[m]{\alpha})/K)$ into $\mu_m$, the set of primes $\mathfrak p$ in $\mathbf Z[\zeta_m]$ such that $(\frac{\alpha}{\mathfrak p})_m = \zeta$ has natural density $1/[K(\sqrt[m]{\alpha}):K]$.
In particular, if $x^m - \alpha$ is irreducible over $\mathbf Q(\zeta_m)$ then for each $m$th root of unity $\zeta$, the set of primes $\mathfrak p$ in $\mathbf Z[\zeta_m]$ not dividing $(\alpha)m$ such that $(\frac{\alpha}{\mathfrak p})_m = \zeta$ has natural density $1/m$.
You are asking specifically about primes $\mathfrak p$ in $\mathbf Z[\zeta_m]$ such that $(\frac{\alpha}{\mathfrak p})_m = 1$. That means $\mathfrak p$ splits completely in $\mathbf Q(\zeta_m,\sqrt[m]{\alpha})$ (an unramified prime splits completely in a Galois extension if and only if its Frobenius conjugacy class is trivial). The Chebotarev density theorem implies that the set of primes splitting completely in a Galois extension of degree $n$ has density $1/n$, and in your case $1/n = 1/m$ if and only if $x^m - \alpha$ is irreducible over $\mathbf Q(\zeta_m)$. So you need to determine whether $x^m - (1 - \zeta_m)$ is irreducible in $\mathbf Q(\zeta_m)[x]$, where the parentheses around the constant term just collect terms together; they do not indicate a principal ideal.
The Chebotarev density theorem is almost 100 years old (proved by Chebotarev in the 1920s), but to compute the density of primes with a given Frobenius element in an abelian Galois group we can use instead a weaker result than Chebotarev's density theorem called the Frobenius density theorem (proved in 1880). This result of Frobenius is about the density of primes with Frobenius element lying in something possibly larger than a conjugacy class of a Galois group when the group is not abelian, but for abelian Galois groups, the Frobenius and Chebotarev density theorems become the same statement. Since $K(\sqrt[m]{\alpha})/K$ is an abelian Galois extension, the fact that the density of $\mathfrak p$ such that $(\frac{\alpha}{\mathfrak p})_m$ has a particular value (like $1$) is $1/[K(\sqrt[m]{\alpha}):K]$ could be considered a consequence of the Frobenius density theorem.