This theorem and it's proof from number theory by Sierpinski may help you to find such numbers or filter numbers to increase the efficiency of calculations:
1-Prove there exist arithmetic progressions with arbitrary number of terms, such that it's different terms are perfect powers of natural numbers.
Proof by A. Schnitzel: take $P_k$ as the kth prime number, $a$ as an arbitrary natural number and $P=p_1p_2p_3...p_s$. Due to Chineas remainder theorem for any natural number such as $k\leq s$, there exist natural number $a_k$, such that we have:
$\begin {cases}a_k\equiv 0\bmod \frac P{P_k}\\ a_k \equiv -1 \bmod P_k\end{cases}$
Suppose $Q= 1^{\alpha}1\cdot 2^{\alpha}2\cdot\cdot\cdot S^{\alpha}s$, it is clear that numbers $KQ, (k=1,2,3,...s)$ make an increasing arithmetic progression which contains s terms. From definition of number $a_k,( k=1,2,3, ...s)$ we conclude that $a_k+1 $ and $a_n$ are divisible by $a_n$ and $p_k$ respectively, for $k\neq n$, where $n\leq s$. In this way numbers such as :
$$Q_k=k^{\frac{a_k+1}{P_k}}\prod^s_{n=1}n^{\frac{a_n}{P_k}} ; n\neq k$$
are natural and we can easily search that for $k=1, 2, ...s$ numbers like $kQ$ are powers, greater than unity, of natural numbers.
Update: You may also use this theorem from the same book:
Prove there does not exist arithmetic progression with infinite number of terms, such that it's different terms are perfect powers of natural numbers.
Proof: We must prove that this equivalent theorem that says in every increasing arithmetic progression with infinite terms , all natural numbers, there exist numbers which are not perfect powers of natural numbers.
Suppose $ak+b; (k=0, 1,2,...)$ is a progression with infinite terms.There exist always prime number $p>a+b$. Since $(a, p^2)=1$ , equation $ax-p^2 y=1$ has solutions for natural numbers x and y. We take $k=(p-b)x$, it is obvious that this number and $ak+b=p^2y(p-b)+p$ are natural numbers,(because p>b). hence term $ak+b$ is divisible by p, but not by $p^2$, so it can not be a power of a natural number.
