Here is a silly thing; I am not sure it is an "advantage" (or, for that matter, a disadvantage), but it indicates a difference:
Inside a model $M$ of $\mathsf{ZF}$ there may be "hidden'' models $N$ of $\mathsf{ZF}$. The situation I have in mind is something like the following, which uses the fact that $\mathsf{ZF}$ is not finitely axiomatizable: $\omega^M$ is nonstandard, $n\in M$ is an infinite natural number, $N\in M$, $M\models E\subseteq N\times N$, and $M$ does not think that $(N,E)\models\mathsf{ZF}$ because it fails to satisfy, say, replacement for $\Sigma_n$ formulas, and $n$ is smallest possible. In this situation, $(N,E)$ is actually a model of $\mathsf{ZF}$, but $M$ does not know it.
(Or, rather than $(N,E)$, the structure $(N^*,E^*)$, where $N^*=\{a\in M : M\models a\in N\}$ and, for $a,b\in N^*$, $a\mathrel{E^*}b$ if and only if $M\models a\mathrel{E}b$.)
This situation actually happens, since by the reflection theorem, any finite fragment of $\mathsf{ZF}$ has models, but $M$ could be a model of $\lnot\mathrm{Con}(\mathsf{ZF})$. So, given such an $M$ (which, necessarily, has a nonstandard $\omega$), the finite (from the point of view of $M$) fragment of $\mathsf{ZF}$ where for some infinite (from the outside) $k\in\omega^M$ we only consider replacement for $\Sigma_m$ formulas with $m<k$ has a model $(N,E)$ in $M$. Since $M$ believes that $\mathsf{ZF}$ is inconsistent, there must be a least $n$ (which, necessarily, is infinite from the outside) such that $(N,E)$ does not satisfy replacement for $\Sigma_n$ formulas.
On the other hand, there can be no such "hidden" models of $\mathsf{NBG}$, by finite axiomatizability.
