Are more numbers in this sequence that starts with $2$,$4$,...?
A number is in this sequence if all its factors add up to the same number as the product of the prime of its digits.
so 2 is in this list because $1$+$2$=prime($2$), and $4$ is in this list because $1+2+4$ =Prime($4$)
12 isn't in this sequence because $1+2+3+4+6+12$ is not $=$ to prime($1$)$\times$prime($2$)
** for 10 we would have Prime(1) we ignore 0 so 120 would be prime(1)$\times$prime(2)
all I want to know is there more in this sequence and is there infinity more?
There are no more. Note that since the $9$'th prime is $23$, the sum of primes of the digits of an $n$-digit number is at most $23n$. But the sum of factors of an $n$-digit number is at least $10^{n-1}$ (since the number itself is a factor). For $n \ge 3$, $10^{n-1} > 23 n$, so there can't be any terms with $3$ or more digits. And by a search we find the only ones with $1$ or $2$ digits are $2$ and $4$.
You might do better if you used the product rather than sum of primes of the digits: $2, 4, 148, 484$ would work (and I don't know if there are more; that's all up to $10^6$).