I've got a proof for this below but I'm not exactly sure if the result or the proof is correct or not. Thanks!
By Bezout's Identity, there exists integers $s, t$ such that $ms + nt = d$. Then, $a^{\frac{d}{nm}} = a^{\frac{ms + nt}{nm}} = a^{\frac{t}{n} + \frac{s}{m}} = (a^{\frac{1}{n}})^{n} \cdot (a^{\frac{1}{m}})^{s} \in \mathbb{Q}(a^{\frac{1}{n}}, a^{\frac{1}{m}})$, so $\mathbb{Q}(a^{\frac{d}{nm}}) \subseteq \mathbb{Q}(a^{\frac{1}{n}}, a^{\frac{1}{m}})$ since it is the smallest field containing $a^{\frac{d}{nm}}$ and $\mathbb{Q}$.
Since $d \mid m$, there exists some integer $k > 0$ such that $m = dk$. Then, $a^{\frac{1}{n}} = (a^{\frac{d}{nm}})^{\frac{m}{d}} = (a^{\frac{d}{nm}})^{k} \in \mathbb{Q}(a^{\frac{d}{nm}})$. We can do the same for $a^{\frac{1}{m}}$. So $\mathbb{Q}(a^{\frac{d}{nm}}) \supseteq \mathbb{Q}(a^{\frac{1}{n}}, a^{\frac{1}{m}})$.