Finitely many Supreme Primes?

This is not an answer, just a bit too long to be a comment. I didn't write the code for finding supreme primes, but I think it is simple.

All supreme primes $x$ are of the form:

$$x = \sum_{k=0}^n 10^k + 10^w\times(p-1) = \frac{10^{n+1} - 1}{9} + 10^w\times(p-1) \tag{1}$$

where $p$ is a prime number, and $0\le w \le n$. Therefore you only need to explore varying three parameters: $n,w,p$. Moreover, the search can be restricted so that $n + p$ (digit sum) and $n + 1$ (number of digits) are prime numbers (see comments). Defining $q = n +1$, we have to search pairs of prime numbers $p,q$ such that $p + q - 1$ is also a prime number (see comments). Having found such a pair, search for a $w$ in the range $0\le w\le q - 1$ such that $x$ in (1) is prime.

Just to clarify (there was some confusion in the comments), note that an $x$ of the form (1) may not be a supreme prime; indeed we still need to know that $x$ itself is prime.

Heuristically, a random $n$-digit number has probability approximately $c/n$ of being prime, where $c$ is a positive constant. Since $p$ is $3$ or $5$ or $7$ and there are $n$ possible locations for it, there are $3n$ $n$-digit candidates. So the expected number of primes of this form among the $n$-digit candidates should be approximately constant. This would indicate that there should be infinitely many supreme primes. Of course it is not a proof. It would be very surprising if this could be proven.

Only some extended hints:

The numbers are necessarily of the form $N=\frac{10^n-1}9+a10^k$ with $0\le k<n$ and $a+1$ prime (i.e. $a\in\{1,2,4,6\}$) to meet the product criterion. Of course, $n$ must be prime to meet the length criterion. Next, the digit sum is $n+a$, which rules out $a=1$ except for small cases.

If $a=2$ we must have $n\equiv -1\pmod 6$. Then $N\equiv n-k \pmod 3$, so we need $k\not\equiv -1\pmod 3 $. Also , $\frac{10^n-1}9\equiv 2\pmod 7$, ${}\equiv 9\pmod {13}$, ${}\equiv 11\pmod{37}$, which rules out another fraction $\frac 17$, $\frac 1{13}$, $\frac1{37}$, respectively, of possible $k$ values.

Similarly, if $a=4$ we must have $n\equiv 1\pmod 6$. Then $N\equiv n+k\pmod 3$, so we need $k\not\equiv -1\pmod 3$ again. The argumentation with the primes $7,13,37$ dividing $111111$ applies as well in a similar way.

More generally, for any (small) prime $p>5$ we can obtain restrictions for $n,k\pmod{p-1}$. This makes it easy to construct $N$ which has no small prime divisors (and of course fulfills the sum, product, and length criterion). There's still a long way to go to make $N$ prime, of course.

Heuristically, the chance of $N$ being prime is $\sim \frac1n$; and since we have $\sim n$ choices for $k$, we'd expect a certain more or less constant amount of supreme primes per $n$ ...