First-order sentence involving only a symmetric binary relation with only infinite models It's known that there are sentences of first-order logic which only have infinite models, even if our language consists only of a binary relation $R$. An example of such a sentence is 
$$\forall x \exists y Rxy \wedge \forall x \forall y \forall z ((Rxy \wedge Ryz ) \to Rxz) \wedge \neg \exists x Rxx$$
What I'm curious about is whether there are still sentences with only infinite models when we mandate that our binary relation be symmetric?
To phrase the question formally, consider a first-order language $\mathcal{L} = \{R^{2}\}$ (without equality). Let $\phi$ denote the formula
$$\forall x \forall y (Rxy \to Ryx)$$
Is there a sentence $\psi$ of $\mathcal{L}$ such that $\phi \wedge \psi$ has an infinite model, but no finite models?

To formalise Stephen A. Meigs's suggestion, define:
$$\sigma (x,y) = \forall z (Rxz \leftrightarrow Ryz) \\
\kappa (x) = \forall y_{1} \forall y_{2} \forall y_{3} ((Rxy_{1} \wedge Rxy_{2} \wedge Rxy_{3}) \to (\sigma (y_{1},y_{2}) \vee \sigma (y_{2},y_{3}) \vee \sigma (y_{1},y_{3})))  \\
\lambda(x) = \exists y(Rxy \wedge \forall z(Rxz \to \sigma(y,z)))
$$
Then let $\psi$ be the conjunction of:
$$
\forall x \kappa (x) \\
\exists x (\lambda (x) \wedge \forall y (\lambda (y) \to \sigma(y,x))) \\
\forall x \forall y \neg (\sigma(x,y) \wedge Rxy)
$$
