If you know that every positive natural number can be written as $a^2 b$ with $b$ being the product of distinct primes (not dividing $a$), just assume that there are just $M$ primes $p_1,p_2,\ldots,p_M$ and consider
$$ \sum_{n=1}^{N}\frac{1}{n} \tag{1}$$
for some large $N$. It is well known that such sum behaves like $\log(N)$. On the other hand, every $n\in[1,N]$ can be written as $a^2 b$ with $a\leq \sqrt{N}$ and $b$ being the product of the elements of a (possibly empty) subset of $\{p_1,\ldots,p_M\}$. In particular:
$$ \sum_{n=1}^{N}\frac{1}{n} \leq \sum_{a\leq \sqrt{N}}\frac{1}{a^2} \prod_{k=1}^{M}\left(1+\frac{1}{p_k}\right)\leq \zeta(2)\cdot C_M\tag{2} $$
is bounded by an absolute constant, depending on $M$ only.
But the LHS of $(2)$ is arbitrarily large, hence the set of prime numbers cannot be finite, qed.
Indeed $(2)$ can be used to prove a very weak form of the prime number theorem:
$$ \sum_{\substack{p\text{ prime}\\p\leq N}}\frac{1}{p}\geq \log\log N-O(\log\log\log N).\tag{3}$$
About that, see Mertens' theorems.
Let $a^5+b^5+c^5+d^5=e^5$. Then $(ae)^5+(be)^5+(ce)^5+(de)^5=e^{10}$, and thus $(ae)^{15}+(a^2be^3)^5+(a^2ce^3)^5+(a^2de^3)^5=(ae^2)^{10}$.
Also, note that if $(A,B,C,D,E)$ is a solution of given equation, then $(k^2A, k^6B, k^6C, k^6D, k^3E)$ is also the solution.
Currently we we only know two positive solutions of $a^5+b^5+c^5+d^5=e^5$, which are
\begin{align*}
&27^5+84^5+110^5+133^5=144^5\quad \text{(L. J. Lander, T. R. Parkin, 1966)}\\
&55^5+3183^5+28969^5+85282^5=85359^5\quad \text{(J. Frye, 2004)}
\end{align*}
For the first solution, we have $A=2^4\times 3^5$, $B=2^{14}\times 3^{13}\times 7$, $C=2^{13}\times 3^{12}\times 5\times 11$, $D=2^{12}\times 3^{12}\times 7\times 19$, $E=2^8\times 3^7$. Now we may take $k=2^2\times 3^2$, and this gives your result.
For the second solution, we have $A=3\times 5^2\times 11\times 5693$ (at this point $k=1$ or $5$), $B=3^4\times 5^5\times 11^2\times 1061\times 5693^3$ (at this point $k=1$), and so on, which gives another "minimal" counterexample which is not related to your solution (but which have a nontrivial common factor). This gives
\begin{equation*}
4694745^{15}+5988388578529986097425^5+54501297119520944786775^5+160446671301977466025950^5=400738738455^{10}.
\end{equation*}
See https://www.wolframalpha.com/input?i=4694745%5E15%2B5988388578529986097425%5E5%2B54501297119520944786775%5E5%2B160446671301977466025950%5E5-400738738455%5E10.
Best Answer
Let $p$ be a prime. Henceforth, work modulo $p$. Consider $ P = \{ s \pmod{p} | s \in S \}$.
Hint: Show that $P$ is either a singleton, or the entire residue class.
Looking at $P$ is very natural, and the result is reasonably easy to guess by looking at small cases of $p$.
EG For $ p = 7$, try the cases where we have seeds $\{1\}, \{2\}, \{3\}$ and see what you get. These cases are somewhat distinct, have a lot to offer, and might suggest ways to prove the hint.
This takes some work. If you're stuck, show what you've tried.
Corollary: For any $ a \in S$, given any prime $p$ that doesn't divide $ a^2 - a + 1 $, then $ a^2 + 1 \not \equiv a \pmod{p}$, so $P$ has at least 2 elements thus is the entire residue class, and thus there exists some element of $S$ that is a multiple of $p$.
Hence, the set of primes that does not divide any element of $S$, is a subset of the prime factors of $a^2 - a + 1$ (and thus is finite).
Notes: