Solution 1:

This is not so few. If we take the probability that a number $n$ is prime to be $1/\log(n)$, we would expect $\sum \frac 1{n \log (10)}\approx \frac 1{2.3}(\log(n)+\gamma)$ of them out to $n$. If we do the sum from $2$ to $10^4$ it is $3.8$ while we have $3$ and if we do it out to $16208$ we should have $4$ and we do. The sum diverges, so we should have infinitely many, but it diverges very slowly.

Solution 2:

See https://oeis.org/A005042

There you can find out that Martin Gardner first asked Neil Sloane this question.

The next term consists of the first 16208 digits.

A naive probabilistic argument suggests that the sequence is infinite.