For all $i$, we set $b_{i}=a_{i} / d$, so that $\operatorname{PGCD}\left(b_{1}, \ldots, b_{n}\right)=1$, and we also set $B=b_{1} \times \ldots \times b_{n}$ and $\Delta=\operatorname{PGCD}\left(b_{1}^{n}+B, b_{2}^{n}+B, \ldots, b_{n}^{n}+B\right)$. Then $\operatorname{PGCD}\left(a_{1}^{n}+A, a_{2}^{n}+A, \ldots, a_{n}^{n}+A\right)=d^{n} \times \Delta$, and it remains to prove that $\Delta \leqslant 2$.
We then consider a potential prime factor $p$ of $\Delta$. If $p$ divides one of the integers $b_{i}$, then it also divides $B$, so it divides each integer $b_{j}$, which is absurd. Therefore, $p$ does not divide any of the integers $b_{i}$, and does not divide $B$ either. This shows that $\Delta$ is coprime with $B$. Since $b_{1}^{n} \equiv b_{2}^{n} \equiv \ldots \equiv b_{n}^{n}=-B(\bmod \Delta)$, then $B^{n} \equiv b_{1}^{n} \times \ldots \times b_{n}^{n} \equiv(-1)^{n} B^{n} \equiv-B^{n}$ $(\bmod \Delta)$. Therefore, $\Delta$ divides $2 B^{n}$. We conclude that $\Delta$ divides 2, which completes the proof.