I came up with what I think is a nice elementary proof, and I just want to double-check there are no mistakes:
Proof that there is no simple group $G$ of order $1080=2^33^35$.
Suppose $G$ is simple.
The candidates for the number of $5$-Sylow subgroups are $6$, $36$, and $216$. $216$ is impossible as the $5$-Sylows would then be in the centers of their normalizers, implying the existence of a normal $5$-complement, by the Burnside transfer theorem. $6$ is impossible as it would imply a non-trivial homomorphism from $G$ to $\frak{S}_6$ (the transitive action of $G$ on the $6$ $5$-Sylows), which is impossible as $|G|>|\frak{S}_6|$ and $G$ is simple. Thus, a $5$-Sylow $P_5$ has a normalizer $N_5$ of order $\frac{1080}{36}=30$. Since $P_5$ is cyclic of order $5$, ${\rm Aut}(P_5)$ is cyclic of order $4$. Thus, the $3$-elements in $N_5$ commute with the $5$-elements. Note that in $N_5$, the resulting cyclic subgroup of order $15$ has index $2$ and thus is normal. The (Sylow) $3$-subgroup of $N_5$ is thus normal (and unique) as it is characteristic in a normal subgroup.
Now consider a $3$-element $z$ that centralizes a $5$-element. (These exist in $N_5$.) This exists in some $3$-Sylow, and in a group of order $27$ every centralizer has order at least $9$. ($p$-groups have non-trivial centers, so non-central elements are centralized by themselves and and non-trivial element of the center.) Thus, $|Z(z)|$ is divisible by $45$. However, the $5$-Sylow can't be normal in $Z(z)$ as its normalizer in $G$ only has order $30$. The only candidates for the number of $5$-Sylows in $Z(z)$ is $6$, since if it were $36$, it would be a normal subgroup (the set of all $p$-Sylows is closed under conjugation). Thus, each $3$-element $z$ that centralizes a $P_5$ centralizes $6$ of them.
But we saw above that each $P_5$ is normalized by a unique $3$-subgroup $\langle z\rangle$ of order $3$. We now see that each such $\langle z\rangle $ normalizes $6$ $5$-Sylows. Since there are $36$ $5$-Sylows, there are $6$ such $\langle z\rangle $ (one-sixth as many as there are $5$-Sylow subgroups), and $G$ acts non-trivially on them by conjugation, and thus has a non-trivial homomorphism to $\frak{S}_6$. Thus $G$ cannot be simple.