Just as when Euclid's proof is erroneously reported (by Dirichlet, G. H. Hardy, and others) to be a proof by contradiction, your proof is more complicated than it needs to be. Instead of supposing that only finitely many exist, just say you have some set of finitely many, then go through your argument to show you can extend it to a larger finite set.
Other than that I think you're OK.
All prime divisors of $\frac{(pr)^{p}-1}{pr-1}$ for any positive integer $r$ are of the form $pk+1$ (proof below).
Now assume for contradiction there are finitely many primes $\{p_1,p_2,\ldots, p_t\}$ of the form $pk+1$. Let $r=p_1p_2\cdots p_t$.
Then all prime divisors of $\frac{(pr)^{p}-1}{pr-1}$ are of the form $pk
+1$ and coprime to $p_1,p_2,\ldots, p_t$. Contradiction.
Proof: Define $\text{ord}_n(a)$ to be the least positive integer $m$ such that $a^m\equiv 1\pmod{n}$.
Lemma: if $x^t\equiv 1\pmod{n}$, then $\text{ord}_n(x)\mid t$. Proof: if $t=\text{ord}_n(x)l+r,\, 0<r<\text{ord}_n(x)$, then $1\equiv x^t\equiv \left(x^{\text{ord}_n(x)}\right)^lx^r\equiv x^r\pmod{n}$, contradiction. q.e.d.
Let $q$ be a prime divisor of $\frac{(pr)^{p}-1}{pr-1}$. Then $\text{ord}_q(pr)\in\{1,p\}$ and $\gcd(q,pr)=1$. For contradiction, assume $q\mid pr-1$.
$$q\mid (pr)^{p-1}+(pr)^{p-2}+\cdots+1\equiv 1^{p-1}+1^{p-2}+\cdots + 1\equiv p\not\equiv 0\pmod{q},$$
contradiction. Therefore $\text{ord}_q(pr)=p$, so by Fermat's Little Theorem and the lemma $p\mid q-1$. Q.E.D
Best Answer
I'm sure this isn't fastest, but it's super-linear. It is known that for $k$ sufficiently large, there is always a prime between $k^3$ and $(k+1)^3$. Now consider the following function. If $n = 2^{2m+1}+j$, $0 \le j < 3 \cdot 2^{2m+1}$, then $f(n) = 2^{3m}+j$. Note that $2^{3m} + 3 \cdot 2^{2m+1} > (2^m+1)^3$ for $m \ge 1$. Thus for each sufficiently large $m$, there will always be $0 \le j < 3 \cdot 2^{2m+1}$ for which $f(2^{2m+1}+j)$ is prime.