The number $1212...2121$ with $n$ ones can be represented by $$z_n\ := \frac{2\times 100^n-35}{165}$$ or, substituting $m=2n-1$, by $$z_n\ :=\ \frac{4\times 10^m-7}{33}$$
It is easy to show that $2,3,5,7$ cannot be a factor of such a number. $11$ can be a factor, which can be shown by $11|121$. Obviously, greater prime factors $p$ must fulfil the equation $$4\times 10^m\equiv 7\ (\ mod\ p\ )$$ with some odd number $m$ or equivalent the equation $$2\times 100^n\equiv 35\ (\ mod\ p\ )$$
Is there an easy criterion, for which primes $p$, there is an $n$ (or an odd $m$), such that the equation holds ?
A necessary condition that $p\ge13$ occurs, is that $70$ is a quadratic residue modulo $p$. If $70$ is a quadratic residue modulo $p$ and $ord_{100}(p)=p-1$ additionally holds, $p$ will occur infinite many often.
Not at first glance,
$$z_2=121=11^2$$ $$z_3=12121=17 \cdot 23 \cdot 31$$ $$z_4=1212121=1212121$$ $$z_5=121212121=83 \cdot 577 \cdot 2531$$
Then the next prime is $z_6$. I didn't find any other $z_n$ that are prime with $ n \lt 16$.
Organizing the primes found so far, up to $n=16$, we have,
$11$,$17$,$23$,$31$,$83$,$263$...
The Case of $13$
For some reason, $13$ was the first prime missing from the above list. Using the divisibility rule for $13$ given here. I guess one could prove that, $z_n \mod 13$ can have values of $[1,4,5]$ that repeat in that order and nothing else.
Conclusion
As mentioned in the Wikipedia article, the general test for divisibility of a number is not explicitly known. Since the sequence of numbers you present is no different from a collection of random numbers, with respect to prime factorization, there is no reason to expect a closed form or "easy" solution to the problem.