As Sridhar already explained, Lévy–Montague Reflection is a theorem scheme and not a single theorem which resolves the apparent contradiction, but here are a few additional cool facts.

First, note that ZFC is not finitely axiomatizable (otherwise we would indeed have a contradiction) but there is a recursive listing of the axioms of ZFC. Let's fix such a listing $\phi_0$,$\phi_1$,$\phi_2$,... If $M$ is a model of ZFC, then either $M$ is an $\omega$-model (i.e. the finite ordinals of $M$ are truly finite) or it is not (i.e. $M$ has some nonstandard finite ordinals). Let's see what happens in each case.

Suppose first that $M$ is an $\omega$-model. The recursive listing $\phi_0$,$\phi_1$,$\phi_2$,... exists in $M$ and, by Lévy&ndash;Montague, people living in $M$ believe that $\{\phi_0,\ldots,\phi_n\}$ has a model for each $n < \omega$. Since people living in $M$ also believe in the Compactness Theorem, they also believe that there is a model of ZFC. This is surprising, but note that the hypothesis that $M$ is an $\omega$-model is essential since without it we there is no reason for $M$'s notion of finite to agree with ours.  This is where your initial reasoning strayed, you naturally assumed that every model of ZFC was an $\omega$-model.

Suppose now that $M$ is not an $\omega$-model. The recursive listing $\phi_0$,$\phi_1$,$\phi_2$,... makes sense in $M$, but since $M$ has nonstandard finite ordinals this listing continues beyond the true $\omega$ and people who live in $M$ believe that these nonstandard $\phi_N$'s are real axioms of ZFC! By Lévy&ndash;Montague, $M$ believes that $\{\phi_0,\ldots,\phi_n\}$ has a model for every *standard* $n$, but since Lévy&ndash;Montague Reflection doesn't say anything about nonstandard axioms, there may be some nonstandard finite ordinal $N$ in $M$ such that people living in $M$ do not believe that the nonstandard finite set $\{\phi_0,\ldots,\phi_N\}$ has a model.

Now here is a funny thing that was pointed out by Joel David Hamkins [in answer to another question](http://mathoverflow.net/questions/15685/is-it-necessary-that-model-of-theory-is-a-set/15713#15713). Suppose $M$ is a model of ZFC + ¬Con(ZFC). Since people in $M$ believe that their finite ordinals are wellordered, there must be a first finite ordinal $N$ in $M$ such that $\{\phi_0,\ldots,\phi_N\}$ has no model in $M$. This $N$ must be nonstandard finite ordinal, and so must its predecessor $N-1$. By minimality of $N$, people in $M$ believe that $\{\phi_0,\ldots,\phi_{N-1}\}$ does have a model. Let $M'$ be such a model. Note that $M' \models \phi_n$ for every *standard* axiom $\phi_n$ since $n < N-1$. Therefore, although people living in $M$ certainly don't believe it, this $M'$ is in fact a model of ZFC!!!

Thus, Lévy&ndash;Montague Reflection does imply that every model of ZFC contains another model of ZFC, but the models are not necessarily aware of that fact...