Ec primes dividing ec numbers

Question

A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$.

In this question ec numbers are introduced, formed by the concatenation of two consecutive Mersenne numbers ($157$ for example is denoted by $ec(4)$).
The ec prime $ec(7)=12763$ divides ec numbers $ec(7717)$, $ec(14259)$, $ec(15906)$,...

Does $ec(7)$ divide an infinite number of ec-numbers?

Is $255127$ the largest ec prime dividing at least one ec number besides itself?

Answer 1

The ec-prime $ec(8)$ divides $ec(k)$ for the following exponents up to $10^7$

284274 1129738 1189846 1214317 1301821 1362842 1445186 1795733 1853089 2203032 2
237654 2267753 3055770 3080516 3532082 3624320 3842054 4653541 4839828 5220495 5
436726 5444103 5828733 5956001 6144125 6432347 6821804 7135640 7173850 7458223 7
513523 7690720 7979828 8006289 8010227 8162195 8195920 8255472 8412247 8449267 8
590602 8936597 9571824 9625677 9853929

I do not know how to prove it, but both $ec(7)$ and $ec(8)$ should divide infinite many ec-numbers. For example $(2^{n+1}-1)\cdot 10^m+2^n-1$ is divisible by $ec(7)$ , if $n$ is of the form $12762k+81$ and $m$ of the form $709l+1$ (but not only then!) . And $(2^{n+1}-1)\cdot 10^m+2^n-1$ is divisible by $ec(8)$ , if $n$ is of the form $42521k+1$ and $m$ of the form $85042l+31514$ (but not only then!). I do not know whether even larger ec-primes divide some ec-numbers.