It is well known, as Fraïssé's theorem, that for a finite relational signature $\sigma$ and two $\sigma$-models $\mathfrak{A}$, $\mathfrak{B}$, $\mathfrak{A}$ and $\mathfrak{B}$ are elementary equivalent over the first-order language $\mathrm{FO}(\sigma)$ iff $\mathfrak{A}\simeq_\mathrm{fin}\mathfrak{B}$, that is to say that they are finitely isomorphic via the back-and-forth method.
What would be a counterexample for $\mathfrak{A}\equiv\mathfrak{B}$ iff $\mathfrak{A}\simeq_\mathrm{fin}\mathfrak{B}$ where $\mathfrak{A},\mathfrak{B}$ are structures over an infinite relational vocabulary.
Consider the relational signature $\sigma = \langle P_i \mid i \in \omega \rangle$. Now consider the structures $\mathfrak A = \langle \mathcal P(\omega), P_i^{\mathfrak A}\rangle$ where $P_i^{\mathfrak A}(X) \iff i \in X$, and $\mathfrak B = \langle \mathcal P_{\mathrm{fin}}(\omega) \times \mathbb R, P^{\mathfrak B}_i\rangle$ where $P_i^{\mathfrak B}((X, r)) \iff i \in X$. Then the structures are elementarily equivalent: each sentence only contains finitely many of the relational symbols $P_i$, and if we simply ignore all other relational symbols and their interpretations, then the structures are even isomorphic. However, they are not finitely isomorphic: no point in $\mathfrak B$ is isomorphic to $\omega \in \mathfrak A$, since at $\omega \in \mathfrak A$ all relational symbols are true, while that is true of no point in $\mathfrak B$.