$\bullet \textbf{ Explanation}$
A "substrime" is a prime in which all substrings are prime. The number $373$ is an example since $3,7,37,73,$ and $373$ are each prime numbers. In decimal (base-10) the only substrimes are: $$2,3,5,7,23,37,53,73,373.$$ See: https://oeis.org/A085823
But we can look at substrimes in any base. For example in base-5 the substrimes are $$2,3,23,32,232.$$
There are $103$ substrimes in base-24 the largest of which is $$266003=(19)(5)(19)(11)|_{24}.$$
We computed these up to base-100 and found base-90 had the most; $1455$ in total (live code: https://repl.it/@onnomc/substrimes).
$\bullet \textbf{ Question}$
Does any base have infinitely many substrimes?
$\bullet \textbf{ Attempt}$
Every substrime of $n$ digits can be thought of as two substrimes on $n-1$ digits such that cutting off the leftmost digit of one and the rightmost digit of the other give a substrime of length $n-2$. For example, in base $28$ the substrime $$(11)(3)(23)(3)|_{28}=244471$$ has $4$ digits. It can be viewed as the substrimes $$(11)(3)(23)|_{28}=8731\quad\text{and}\quad(3)(23)(3)|_{28}=2999$$ which share a common substrime $(3)(23)|_{28}=107$ obtained after truncation.
So let "$S_b(n)$" mean "the amount of substrimes with $n$ digits in base $b$". From the previous bit we obtain two inequalities $$(1)\quad S_b(n)\le S_b(1)\cdot S_b(n-1)$$ $$(2)\quad S_b(n)\le S_b(n-1)^2$$ which paired with the prime number theorem gives us heuristics $$(1)\quad S_b(n) \approx \frac{S_b(1)\cdot S_b(n-1)}{n\log b}=C_b\frac{S_b(n-1)}{n}$$ $$(2)\quad S_b(n)\approx C_b'\frac{S_b(n-1)^2}{n}$$
Thus we would expect $S_b(n)$ to hit zero eventually since the numerator grows at best like O$(n^2)$ but the denominator grows like O$(n!)$.
But of course, this is hueristic. It seems also highly composite bases have more substrimes. For example, the record holder bases with the most susbstrimes are: $$3,4,5,7,8,12,18,24,30,42,48,54,60,90,...$$
Here is a chart:
I was unable to determine why composite bases have more substrimes.