Are there infinitely many primes of the form $6^{2n}+1$ or only finitely many? Does anyone know whether there are only finitely many of primes of the form $6^{2n}+1$, where $n$ zero or any natural number?
 A: It is not known whether there are infinitely many primes of the form $a^2 + 1.$ Your numbers are the subset with $a = 6^n.$ So it is not known that your set is infinite. The strongest results are in two variables, the well-known result that infinitely many primes are of the form $x^2 + y^2$ (indeed all primes $p \equiv 1 \pmod 4$) and the result of Iwaniec and Friedlander that there are infinitely many primes of the form $x^2 + y^4.$  
A: I'm pretty sure the in-text question is "no," which means the title question is pretty tough to answer cleanly.  But just for the sake of mentioning it, it's easy to check that $n$ would have to be a power of 2 for this quantity to prime:  Indeed, if we re-write it as $36^n+1$, then the argument is verbatim the same as for Fermat primes:  If $n$ were not a power of two, it would admit a proper factorization $n=ab$ with (say) $b$ odd.  Then $36^a+1\mid 36^{ab}+1$, contradicting primeness.  
So now you're looking for primes of the form $36^{2^k}+1$, a problem which doesn't, at least to me, appear any more tractable than the (wide open) analogous problem with base 2, the problem of finding the Fermat primes (taking the form $2^{2^k}+1$).  In fact, there is some literature on the topic -- mostly computational, as primes of this form are rare but algorithmically interesting -- going under the eminently-Googlable name "generalized Fermat number."
Finally, let me just note that via direct computation (SAGE), $36^2+1$ is prime, and $36^{2^k}+1$ is composited for $2\leq k\leq 6$.
