What is the smallest size of a set $S$ with some extra conditions such that $S$ contains an $n$-th power residue for each prime $p$? This post is inspired by this one.  I have two related questions.

Definition.  Let $n$ be a positive integer.  For an integer $m$ and $a$, we say that $a$ is an $n$-th power residue modulo $m$ if $$x^n\equiv a\pmod{m}$$ has a solution $x\in\mathbb{Z}$.  A subset $S\subseteq \mathbb{Z}$ is said to be $n$-th power saturated if, for each prime natural number $p$, $S$ contains an $n$-th power residue modulo $p$.

Examples.  The set of all prime natural numbers itself is $n$-th power saturated for every positive integer $n$.  The set $\{2,3\}$ is not $n$-th power saturated for $n\in\{2,3,4\}$ (i.e., it is not quadratic-saturated, cubic-saturated, or quartic-saturated).

Question 1.  For each positive integer $n$, what is the smallest cardinality of an $n$-th power saturated subset $S$ of the set of positive integers $\mathbb{Z}_{>0}$ such that 
(a) $S$ contains no $n$-th perfect powers?  (Let the answer be $A_n$.) 
(b) $S$ contains no perfect powers at all?  (Let the answer be $B_n$.) 
(c) $S$ contains only squarefree integers? (Let the answer be $C_n$.)


Question 2.  Does there exist a finite set $S$ containing integers greater than $1$ such that $S$ is $n$-th power saturated for any positive integer $n$?
Update. The answer is no.  See Carl Schildkraut's answer below.

Known Results.  Obviously, $$A_n\leq B_n\leq C_n\,.$$  From this answer, we know that $$C_n\leq \dfrac{n(n+1)}{2}\,.$$  It is known that $$A_2=B_2=C_2=3$$ due to Chebotarev's Theorem.  (I have not seen a proof of this claim about the case $n=2$, so a reference for this would be appreciated.)  From  user760870's comment below, we can see that
$$B_4\leq 6\,,$$
by taking $S=\{2,3,6,12,18,24\}$.  The same user claimed in the same comment that $$A_n=B_n=n+1\text{ if $n$ is prime}\,,$$ with $S=\{2,3,6,12,\ldots,3\cdot 2^{n-1}\}$ as an example (I understand why this choice of $S$ works, but I cannot yet prove that this set $S$ has the lowest possible cardinality).  It is easy, however, to verify that
$$A_{8}=1\,,$$
by taking $S=\{16\}$, per the comment by user760870.  Consequently,
$$A_{2^k}=1\text{ for all }k=3,4,5,\ldots\,,$$
by taking $S=\left\{2^{2^{k-1}}\right\}$.

  This is a proof that $A_8=1$, where $S=\{16\}$ works.  Note that $$x^8-16=(x^2-2)(x^2+2)(x^4+4)\,.$$   - If $p=2$, then $x=0$ is a trivial solution.  - If $p\equiv 1 \pmod{8}$ or $p\equiv 7\pmod{8}$, then $x^2-2\equiv 0\pmod{p}$ has a solution $x\in\mathbb{Z}$.   - If $p\equiv 3\pmod{8}$, then $x^2+2\equiv 0\pmod{p}$ has a solution $x\in\mathbb{Z}$.   - If $p\equiv 5\pmod{8}$, then let $t\in\mathbb{Z}$ satisfy $t^2+1\equiv0\pmod{p}$, and note that $2t$ is a quadratic residue modulo $p$ (since both $2$ and $t$ are not).  Therefore, $x^2-2t\equiv 0\pmod{p}$ has a solution $x\in\mathbb{Z}$, whence $$x^4+4\equiv (x^2-2t)(x^2+2t)\pmod{p}$$ implies that $x^4+4\equiv0\pmod{p}$ has a solution $x\in\mathbb{Z}$.   From this result, we can then conclude that $A_{2^k}=1$ with $S=\left\{2^{2^{k-1}}\right\}$, where $k\geq 3$ is a positive integer.  This is because $x^{2^k}-2^{2^{k-1}}$ is divisible by $x^8-16$.  In fact, $$x^{2^k}-2^{2^{k-1}}=(x^2-2)\,\prod_{j=1}^{k-1}\,\left(x^{2^j}+2^{2^{j-1}}\right)=(x^8-16)\,\prod_{j=3}^{k-1}\,\left(x^{2^j}+2^{2^{j-1}}\right)\,.$$

 A: Here's an elementary answer to question 2.
We claim that no set of positive integers, each of which is strictly between $1$ and $k$ is $(p-1)$-th power saturated for any prime $p>k$. Indeed, for such a set $S$ to be $(p-1)$-th power saturated, it must contain an integer $n$ so that
$$n\equiv x^{p-1}\bmod p.$$
However, by Fermat's Little Theorem, $x^{p-1}\in\{0,1\}\bmod p$, and no element of $S$ can be $0$ or $1$ modulo $p$, since all are less than $p$ and greater than $1$.

Here's a proof that $A_p=p+1$ for prime $p$.
Lemma. Let a subspace $V$ of $\mathbb{F}_p^k$ satisfy that, for each $1\leq i\leq k$, there exist some $x\in V$ for which $x_i\neq 0$. Then, as long as $k\leq p$, there exists some $x\in V$ so that $x_i\neq 0$ for each $1\leq i\leq k$.
Proof. We use the probabilistic method. Pick an $x\in V$ uniformly at random. For each $i$, since $\{y_i \colon y\in V\}$ is not $\{0\}$, the values $y_i$ range uniformly at random throughout $\mathbb{F}_p$, and so
$$\operatorname{Pr}(x_i=0)=\frac1p.$$
As a result,
$$\operatorname{Pr}(\text{any } x_i=0)\leq \sum_{i=1}^k \operatorname{Pr}(x_i=0)=\frac kp.$$
This is enough as long as $k<p$; if $k=p$, we only need the $\leq$ above to be strict. It is, since the zero vector is counted once on the left and $k$ times on the right.

Now, assume that $A_p\leq p$, so there exists a set $\{m_1,\dots,m_p\}$ of positive integers so that, for each prime $q$, some $m_i$ is a $p$-th power. Let $S$ be the set of all primes that divide $\prod m_i$, and associate with each $m_i$ a vector $v_i\in \mathbb{F}_p^S$ so that $(v_i)_r$ is the exponent of $r$ in the prime factorization of $m_i$, when taken modulo $r$.
We claim that there exists a vector $w$ so that $v_i\cdot w\neq 0$ for each $1\leq i\leq p$. Firstly, let $S'\subset S$ be a set of primes of size $\leq p$ so that, for each $i$, $(v_i)_r\neq 0$ for some $r\in S'$; such a prime $r\in S$ exists for each $i$ since no $m_i$ is a perfect $p$th power. We will set $w$ to be $0$ on $S\setminus S'$.
Now, we have $p$ vectors $v_1',\dots,v_p'$ in $\mathbb{F}_p^{S'}$. Let $V\subset \mathbb{F}_p^p$ consist of all vectors of the form
$$\begin{bmatrix}v_1'\cdot w' \\ v_2'\cdot w' \\ \vdots \\v_p'\cdot w'\end{bmatrix}$$
for all $w'\in\left(\mathbb{F}_p^{S'}\right)^\vee.$ We see that for each $i$, since $v_i'$ is not the zero vector, there exists some $w'$ so that $v_i'\cdot w'$ is not $0$, and so $V$ satisfies the conditions of our lemma, whence we can find some $w'$ so that $v_i'\cdot w'\neq 0$ for all $i$, whence $v_i\cdot w \neq 0$ for all $i$, as desired.
Now, we find some large prime $q$ so that no $m_i$ is an $p$th power modulo $q$. We first pick $q$ to be $1\bmod p$ and define a morphism $f:\mathbb{F}_q^\times\to\mathbb{F}_p^+$ by sending some generator of the first to a generator of the second. In particular, under this map, $f(x)=0$ if and only if $x$ is a $p$th power modulo $q$.
We now wish that, for each $r\in S$, $f(r)=w_r$. If this is true, then
$$f(m_i)=f\left(\prod_{r\in S}r^{\nu_r(m_i)}\right)=\sum_{r\in S}\nu_r(m_i)w_r=\sum_{r\in S}(v_i)_rw_r\neq 0,$$
as desired. So, we want, fixing some $p$th root of unity $\zeta$, for
$$\left(\frac rq\right)_p=\zeta^w_r$$
for all $r\in S$ (using the power residue symbol). Eisenstein reciprocity tells us that this is the same as
$$\left(\frac qr\right)_p = \zeta^w_r.$$
This is simply a condition on $q\bmod r$, so by the Chinese Remainder Theorem and Dirichlet's Theorem we can find some valid $q$.
(I'm not sure this final argument with Eisenstein reciprocity is completely right, but the general principle should work.)

For completeness' sake, I'll supply a proof that user760870's stated set works. Set
$$S=\{2\}\cup\{3\cdot 2^i\colon 0\leq i\leq p-1\}$$
to be a set of $p$ positive integers. Assume none of them is a perfect $p$th power modulo some prime $q$, and note that $q\neq 2,3$. Let $g$ be a primitive root modulo $q$, let $h=g^{(q-1)/p}$, and let $k_2$ and $k_3$ be so that $g^{k_i}\equiv i\bmod q$. We have that none of $g^{ak_2+bk_3}$ is a perfect $p$th power modulo $q$ for $(a,b)=(1,0)$ or $b=1$; this implies that
$$h^{ak_2+bk_3}\neq 1$$
for each of these $(a,b)$. In particular,
$$ak_2+bk_3\not\equiv 0\bmod p$$
for any of these $(a,b)$. This means that $k_2\not\equiv 0\bmod p$, so the multiples of $k_2$ form a complete residue system $\bmod p$. This means that $ak_2\equiv -k_3$ for some $a$, a contradiction.
