If for some suitable positive integers $a,b,m, (a,b)=m$, then $\forall i \in \mathbb{Z+}, (a^i,b^i)=m$
Please vet the proof.
Let, for some suitable integers k,l, $a=km, b=lm$. Then $a^i=k^im^i, b^i = l^im^i$.
So, still the $\gcd$ of $(k^im^i,l^im^i)$ is $m$, as $(k,l)=1 \implies(k^i,l^i)=1$.
-- Change in proof due to comments:
So, still the $\gcd$ of $(k^im^i,l^im^i)$ is the highest power of $m$, i.e. $m^i$ as $(k,l)=1 \implies(k^i,l^i)=1$.
If $a=p_1^{\alpha_1}\ldots p_k^{\alpha_k}$ and $b=p_1^{\beta_1}\ldots p_k^{\beta_k}$, then $(a,b)=p_1^{\min\{\alpha_1,\beta_1\}}\ldots p_k^{\min\{\alpha_k,\beta_k\}}$
What then, can you say about $(a^i,b^i)$?
Hint: $\min\{rx,ry\}=r\min\{x,y\}$ for all $r>0$.