You can prove this relatively easily for $m=2$.

Chose $A_1, A_2$ so that there exist infinitely many primes with $q_1q_2$ a quadratic residue and $-A_1A_2$ a non-residue mod $p$. This can be done using quadratic reciprocity and gives the primes lying in an arithmetic progression. (For example if $p\equiv -1 \bmod 24$ then 2 is a residue and -3 is a non residue mod $p$.)

Then $p\mid(A_1q_1^n+A_2q_2^n)$ means that $A_1q_1^n+A_2q_2^n\equiv0 \bmod p$ or $A_1^2q_1^{2n}+A_1A_2{(q_1q_2)}^n\equiv0 \bmod p$ which cannot be true if $-A_1A_2$ is a non residue and $q_1q_2$ is a residue.

Hence all such $p$ cannot divide $(A_1q_1^n+A_2q_2^n)$.