Are there infinite many integer $n\ge 0$ such that $10^{2^n}+1$ prime numbers? It is clear to see that 11 and 101 are primes which sum of digit is 2. I wonder are there more or infinte many of such prime.
At first, I was think of the number $10^n+1$. Soon, I knew that $n\neq km$ for odd $k>1$, otherwise $10^m+1$ is a factor.
So, here is my question:

Are there infinite many integer $n\ge 0$ such that $10^{2^n}+1$ prime numbers?

After a few minutes: I found that if $n=2$, $10^{2^n}+1=10001=73\times137$, not a prime;
if $n=3$, $10^{2^n}+1=17\times5882353$, not a prime; $n=4$, $10^{2^n}+1=353\times449\times641\times1409\times69857$, not a prime. 
Now I wonder if 11 and 101 are the only two primes with this property.

Since no one else has mentioned it:
Standard heuristics in number theory suggest that there are only finitely many primes of the form $(2k)^{2^n}+1$ for any integer $k>0.$ The probability that a random number around $(2k)^{2^n}+1$ is prime is roughly $1/(2^n\log(2k))$; if you take into account the congruence conditions for such numbers and treat the chance that such a number is prime as a random variable, then the expectation is $C_k/2^n$ and the sum over these values converges.
If you sum this 'probability' over $n\ge24$ the expected number of primes of this form is less than 0.000001.
