Primes with prime sum of digits
This is sequence A046704 in OEIS, and according to the web page, this was proven to be infinite.
There is also this paper, Theorem 1 that implies there is an infinite number of such primes and for every large enough prime p, there is a prime whose digits sum is equals to p.
Proofs are not elementary.
Firstly, I suspect that there are infinitely many primes whose digit sums are also primes. Secondly, I would naively expect that we don't know how to prove this.
However, it is possible to show a heuristic that indicates that this should be true. I show that now.
We consider numbers of the form
$$10^n + \sum_{i = 0}^{n-1} \nu_i 10^i,\tag{1}$$
where $\nu_i$ is either $0$ or $1$. If exactly $k$ values of $\nu_i$ are $1$, then clearly the digit sum of such an integer is $k+1$.
On average, a random integer of size up to $10^n$ has a probability of being prime that is about $$ \frac{1}{n \log 10}.$$ Fix $k+1 = p$ some prime relatively near $n/2$ (really, any prime that's not $2,3$ or larger than $n-3$), and consider integers of the form $(1)$ whose digit sum is $p$. There are
$$ {n \choose k } \gg n^2$$
such integers up to size $10^n + 10^{n-1}$. If we assume that each is prime with probability $1/n\log 10$ (this is where we truly enter heuristic territory), then we expect much more than $n$ integers of this form with digit sum $k+1 = p$.
Thus heuristically, I would expect not only that there are infinitely many primes whose digit sums are primes, but that for any $N$, there exists a prime $q$ which has at least $N$ other primes whose digit sums are all $q$.
Note that the assumption that integers of the form $(1)$ have a truly random probability of being prime isn't true in general. For instance, if we happen to want to find a prime whose digits sum to $3$, we will fail (with the exception of $3$, of course). This is because any number whose digits are divisible by $3$ is itself divisible by $3$ (when written in base 10. Similar things are true for primes dividing $b-1$ in base $b$).
But avoiding $3$, I would not be surprised if this heuristic help closer to truth. And it only consider integers containing only $0$s and $1$s in their decimal expansion.