There are indeed no such bipartite graphs. In fact:

**Proposition.** Let $G=(X,Y,E)$ be a bipartite graph with $n:=|X|=|Y|$ in which the degree of each vertex is at least $2$ and at most $3$. Then, $G$ has either no perfect matching or at least two perfect matchings.

**Observation.** Under the hypotheses of the proposition, the number of degree-$2$ vertices is the same in $X$ and $Y$. (Count the number of edges in $G$ as $\sum_{x\in X}d(x)$ or as $\sum_{y\in Y}d(y)$, and equate the two expressions.)

*Proof of the proposition.* We use induction on $n$. The base is when $n=2$, in which case the claim is trivial, because the only bipartite graph satisfying the condition is the cycle of length $4$.

Let $n>2$, and suppose, for the sake of getting a contradiction, that $G$ has exactly one perfect matching. Then, $G$ satisfies Hall's condition, and moreover, there is a set $\varnothing\neq A\subsetneq X$ which is *critical*, in the sense that $|N(A)|=|A|$, where $N(A)$ is the set of neighbors of the vertices in $A$. Namely, if no critical set exists, then Hall's condition remains satisfied even if we remove a single edge from the graph (in particular, one of the edges of the perfect matching), and this contradicts the uniqueness of the perfect matching.

Note that if $A$ is a critical set, then the vertices of $A$ are necessarily matched with $N(A)$, and the vertices in $X\setminus A$ are necessarily matched with $Y\setminus N(A)$. Moreover, every critical set in our graph has at least two vertices.

Let $A$ be a *minimal* critical set. Let $k$ denote the number of degree-$2$ vertices in $A$, and $\ell$ the number of degree-$2$ vertices in $N(A)$. Again, a counting argument shows that $k\geq\ell$, and there are exactly $k-\ell$ vertices between $N(A)$ and $X\setminus A$.

Let $Q$ denote the set of vertices in $N(A)$ that have at least one neighbor in $X\setminus A$, that is $Q:=N(A)\cap N(X\setminus A)$. Note that for every $y\in Q$, $d(y)=3$ and there is exactly one edge between $y$ and $X\setminus A$. Indeed, if this is not the case, then $y$ is connected to exactly one vertex $x\in A$. But then $A\setminus\{x\}$ will also be a critical set, contradicting the minimality of $A$.

It follows that in the subgraph $G'$ which is induced on $A\cup N(A)$, the degree of each vertex is at least $2$ and at most $3$. Therefore, by the induction hypothesis, $G'$ has either no perfect matching or at least two perfect matchings. In either case, this leads to a contradiction because we already know that $A$ can be matched with $N(A)$ and $X\setminus A$ can be matched with $Y\setminus N(A)$. $\square$

thosevertices, there are two perfect matchings.... (What is the source of this problem?) $\endgroup$