The fact that there are arbitrarily long sequences of consecutive numbers without prime number is well known, the proof is easy and goes like this:
let $n\ge 2$, then the number $n!+k$ is greater then $k$ and divisible by $k$ for $k=2,\dots,n$, so it is composite, so we have at least $n-1$ consecutive composite numbers.
I think that there are arbitrarily long sequences of consecutive numbers without prime numbers and without powers of primes, probably also without any perfect powers.
Examples:
$17,18,19,20,21,22,23,24$ -- 8 consecutive numbers without any perfect power,
$37,38,\dots,48$ -- 12 consecutive numbers without any perfect power,
$50,51,\dots,63$ -- 13 consecutive numbers without any perfect power,
$65,66,\dots,80$ -- 16 consecutive numbers without any perfect power,
$901,902,\dots,960$ -- 60 consecutive numbers without any perfect power,
$1601,1602,\dots,1680$ -- 80 consecutive numbers without any perfect power.
But I'm not able to prove my conjecture. Is it really true? Any proposals will be appreciated.
Take $n=2^{2k}$, so that $n$ is a square. Moreover it is easy to see that for all the perfect powers $1<x^t\le n$, the exponent $t$ cannot exceed $2k$.
How many are the powers (2nd, 3rd,... $2k$-th) smaller or equal to $n$ ?
The squares are $2^k$. The other powers, with exponents from 3 to $2k$ are clearly much less, but since we are interested in a bound, let's just assume there are at most $2^k$ of each of them.
All in all, up to $n=2^{2k}$ there are thus less than $2k \cdot 2^k$ perfect powers. So, on the average there is less than a perfect power every $2^{2k}/(2k \cdot 2^k) = 2^k/(2k)$ numbers and thus there must be at least a gap larger than that.
Since we can make grow $2^k/(2k)$ as large as we want, gaps are unbounded.