Let $q_1,q_2,\dots,q_n$ be odd primes of the form $3k+2$. Consider the number $N$, where
$$N=3q_1q_2\dots q_n+2.$$
It is clear that none of the $q_i$ divides $N$, and that $3$ does not divide $N$.
Since $N$ is odd and greater than $1$, it is a product of one or more odd primes. We will show that at least one of these primes is of the form $3k+2$.
The prime divisors of $N$ cannot be all of the shape $3k+1$. For the product of any number of (not necessarily distinct) primes of the form $3k+1$ is itself of the form $3k+1$. But $N$ is not of the form $3k+1$. So some prime $p$ of the form $3k+2$ divides $N$. We already saw that $p$ cannot be one of $q_,\dots,q_n$. It follows that given any collection $\{q_1,\dots,q_n\}$ of primes of the form $3k+2$, there is a prime $p$ of the same form which is not in the collection. Thus the number of primes of the form $3k+2$ cannot be finite.
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
Actually, that proof can be phrased in a way to avoid contradiction. Let ${\cal P}=\{p_1,...,p_n\}$ be a non empty finite set of primes. Then let $a=1+\prod_{p\in\cal P}p$.
By the usual argument there must be a prime $p\mid a$, $p\notin\cal P$. This proves that any finite set of primes cannot include all primes and so there must be infinitely many.
EDIT: Given the almost-infinite sequence of comments, let me spell out the "usual argument":
QED