A usual prime number is a number greater than $1$ which is not in the form of multiplication of two numbers greater than $1$. We may consider the following natural generalizations:
$p>1$ is $+$ - prime iff there are no $m,n>1$ such that $p=m+n$
$p>1$ is $\times$ - prime iff there are no $m,n>1$ such that $p=m\times n$
$p>1$ is $\hat{}$ - prime (or exponentially prime) iff there are no $m,n>1$ such that $p=m^n$
Remark 1: $2,3$ are the only $+$ - prime numbers and $\times$ - prime numbers are the ordinary primes. Also each $+$ - prime is $\times$ - prime and each $\times$ - prime is $\hat{}$ - prime. Furthermore there are many numbers like $6$, $10$ that are $\hat{}$ - prime but not $\times$ - prime.
Question 1: How many $\hat{}$ - prime numbers are there in the interval $[2,n]$ for a given natural number $n>1$? In the other words, is there any formula to compute this as well as $\times$ - primes?
Question 2: Is there any representation theorem for natural numbers in terms of exponentially prime numbers as well as factorization theorem for $\times$ - primes?
Remark 2: Regarding question 2, note that there are too many exponentially prime numbers so we may expect that there is a simpler representation theorem for all natural numbers with respect to this type of primes. More precisely consider the following form of question 2:
Question 3: Is there any fixed number $n_0\geq 0$ such that for all $m>1$ there exists an exponentially prime number $p$ with $m=p+n_0$ (i.e. $p$ is less than or equal to $m$ and near to $m$ in the sense of $n_0$)? What about $|m-p|=n_0$?
Remark 3: Note that in the case of $m=p+n_0$ in above question, $n_0=0$ doesn't work because we may assume $m$ to be a non-exponentially prime number (e.g. $m=4$). Also $n_0=1$ doesn't work because if we consider $m=9$ then $p$ should be $8$ which is not an exponentially prime number.
These are the non-powers, so there are $x-O(\sqrt x)$ of them up to $x$.
For question 1 this can be computed by inclusion-exclusion. Take $\lfloor x\rfloor$, then subtract the $\lfloor\sqrt x\rfloor$ squares up to $x$, then subtract the $\lfloor\sqrt[3]x\rfloor$ cubes up to $x$, then skip the fourth powers (they're already counted), then subtract the $\lfloor\sqrt[5]x\rfloor$ squares up to $x$, then add in the $\lfloor\sqrt[6]x\rfloor$ squares up to $x$, which were counted as both squares and cubes, etc. Generally numbers with exponent $e$ are given weight $\mu(e).$
There's no obvious answer for question 2 since the numbers are so common. One representation you could choose is as a product of primes, which are all ^-primes.
For question 3 you can have the first variant since it requires that you have only finitely many ^-primes (any power plus $n_0$ would fail). A slightly stronger version of Catalan's conjecture would give you the second version with $n_0=1$ and large enough numbers.