In Enderton's logic, Theorems 26C and 26D, the assumption is that the first order language is finite.
Theorem 26C: For a finite structure $\mathfrak A$ in a finite language, $Th \mathfrak A$ is decidable.
Theorem 26D: For a finite language, $\{\sigma | \sigma \text{ has a finite model} \}$ is effectively enumerable.
Is a finite language a necessary assumption in these two theorems? So long as the universe of some structure is finite, one could build a tree from the finite sentence $\sigma$ and evaluate all the leaves with the values from the universe.
If a sentence of the form $\forall c \phi(c)$, where c referred to constants is allowed, then a finite language would be a necessary assumption. But I believe that $\forall$ can only range over variables.
Note that in order to make sense of computability-flavored questions around infinite languages, we need to think of symbols as being represented by natural numbers in some reasonable way. This is something we can ignore when the language is finite, but with infinite languages it's important.
Theorem $26D$ makes no real use of the finiteness of the language. This is because every sentence can only use finitely many symbols. Really the only way complexity can enter the picture is if the language we're looking at is itself of high complexity, in which case the set of sentences in that language itself (not even thinking about the set of sentences in that language with finite models, yet) will be of high complexity. But this has nothing to do with the finite model property. A bit more precisely, as long as $\Sigma$ is a computable language, the set of $\Sigma$-sentences which have finite models is c.e. - given a computable language $\Sigma$ and a $\Sigma$-sentence $\sigma$, let $\Sigma_0$ be the finite fragment of $\Sigma$ which actually appears in $\sigma$ and search through the finite $\Sigma_0$-structures for one satisfying $\sigma$.
Theorem $26C$, by contrast, absolutely does rely on a finite language. Take for example a language with a unary relation symbol $U_i$ for each $i\in\mathbb{N}$ (note that this language is computable!). For each set $X\subseteq\mathbb{N}$, let $\mathfrak{M}_X$ be the structure whose universe consists of a single element, such that $U_i$ holds on that element iff $i\in X$. Then $Th(\mathfrak{M}_X)$ computes $X$ since we have $$i\in X\iff\mathfrak{M}_X\models\exists xU_i(x).$$ So now just pick some really complicated $X$.
Informally speaking, the infinite language lets us "access" each of the bits of $X$ (using one symbol per bit), and so code the entire complexity of $X$ into the theory of the resulting structure. If we tried to do this with only finitely many $U_i$s, we'd only get finitely much information about $X$, which - regardless of how complicated $X$ is - would be computable. And the fact that every finite structure in a finite language has a computable theory demonstrates that there's no clever way around this.