$\textbf{Question:}$Is there an infinite set of positive integers such that no matter how we choose some elements of this set, their sum is not a perfect power?
$\textbf{My progress:}$I thought of solving an easier version,sum not a perfect $k-th$ power for fixed $k$.
In that case I came up with, $p^{d}$ where $d$ ranges over all positive integers not divisible by $k$ and $p$ some prime.But this idea didn't seem good enough to generalize that is I could do for any finite amount of numbers $k$ but not for all $k$ at the same time.
$\textbf{Edit:}$ I used the above idea.The construction is just: $p^dq^{d+1}$ for two different primes $p$ and $q$ and $d$ ranges over all positive integers.
Thanks to everyone who helped.
For each positive integer $m$, let $a_m=2^{m}3^{m+1}$, and let $A=\{a_1,a_2,a_3,...\}$.
Claim:$\;$No nonempty finite subset of $A$ sums to a perfect power (with positive integer exponent greater than $1$) of a positive integer.
Proof:
Suppose instead that there exists a nonempty finite subset $B=\{b_1,...,b_k\}$ of $A$ such that $$b_1+\cdots+b_k=x^y$$ for some positive integers $x,y$, with $y > 1$.
Our goal is to derive a contradiction.
Without loss of generality, assume the sequence $b_1,...,b_k$ is increasing.
Let $n$ be such that $b_1=a_n=2^{n}3^{n+1}$.
Since $2{\,\mid\,}a_m$ for all $m$, it follows that $2{\,\mid\,}x$, and since $3{\,\mid\,}a_m$ for all $m$, it follows that $3{\,\mid\,}x$.
Let $r$ be such that $2^r{\,{\mid}{\mid}\,}x$, and let $s$ be such that $3^s{\,{\mid}{\mid}\,}x$.
By unique factorization, it follows that $2^{ry}{\,{\mid}{\mid}\,}x^y$ and $3^{sy}{\,{\mid}{\mid}\,}x^y$.
Since all terms of the sequence $b_1,...,b_k$ other than $b_1$ are divisible by $2^{n+1}3^{n+2}$, it follows that $2^n{\,{\mid}{\mid}\,}(b_1+\cdots +b_k)$ and $3^{n+1}{\,{\mid}{\mid}\,}(b_1+\cdots +b_k)$.
But then we must have $n=ry$ and $n+1=sy$,$\;$so $y$ is a common factor of $n$ and $n+1$, contradiction, since $y > 1$ and $\gcd(n,n+1)=1$.
This completes the proof.