The multiplicity two square is the quiver on $[4]$ with arrows $i \to i + 1$ modulo $4$ of multiplicity two. Here is code to plot these quivers:
One can produce infinitely many cluster variables by performing the sequence of mutations at the vertices
\[1 \to 3 \to 2 \to 4 \to 1 \to 3 \to 2 \to 4 \to \cdots.\]Here is code that implements this mutation sequence and displays the resultant cluster seeds (note: gets messy very fast).
[proof=Proof of infinitely many cluster variables] Let $Q$ be our quiver as described above and $(Q, \{x_1, \ldots, x_4\})$ be our initial seed. We will specialize by letting $x_1 = x_3 = x$ and $x_2 = x_4 = y$. By direct computation, one obtains
\[\mu_4 \mu_2 \mu_3 \mu_1(Q, \{x, y, x, y\}) = \left(Q, \left\{\frac{2y^2}{x}, \frac{8y^3}{x^2}, \frac{2y^2}{x}, \frac{8y^3}{x^2}\right\}\right).\]Under this specialization, we define
\[X_n = \frac{2^{n(2n - 1)} y^{2n}}{x^{2n - 1}} \quad\text{ and }\quad Y_n = \frac{2^{n(2n + 1)}y^{2n + 1}}{x^{2n}}.\]for $n \ge 1$ and $X_0 = x$ and $Y_0 = y$. By induction and the formula we computed for $\mu_4 \mu_2 \mu_3 \mu_1(Q, {x, y, x, y})$, one can show that
\[(\mu_4 \mu_2 \mu_3 \mu_1)^n(Q, \{x, y, x, y\}) = (Q, \{X_n, Y_n, X_n, Y_n\})\]for all $n \ge 0$. Thus, $Q$ cannot produce finitely many cluster variables. [/proof]