First we will show that at least one has an infinite number of vertices. In fact, let's show that the branch $B$ containing $(0, 0)$ is infinite. Suppose, for contradiction, that it was not. Let $t$ be the smallest nonnegative integer so that over all vertices $(x, y)$ in $B$, we have $\max(|x|, |y|) \le t.$ We claim then that in the $(2t+3) \times (2t+3)$ square centered at $(0, 0)$ (call this $\omega$) there are more than $\frac{t+1}{2}$ empty squares. If $t \le 1$, this is easily verified. Else, suppose that $t \ge 2.$ By symmetry, WLOG that $(t, y)$ is in $B$, where $y$ is positive. Then, it's clear that the intersections of each of the lines $x = 0, x = 1, x = 2, \cdots, x = t$ with $\omega$ have at least one empty square. Hence, we've shown that $(0, 0)$ is in an infinite branch. Let's now show that there are no other infinite branches. Suppose that $(a, b) \in \mathbb{Z}^2$ is some lattice point which is contained in an infinite branch $B'$ which is not $B,$ where $|a| \ge |b|.$ Color red all lattice points in $B$ and blue all lattice points in $B'.$ Define the tastiness of a point $(x, y) \in \mathbb{Z}^2$ to be $\max(|x|, |y|).$ Since this branch is infinite, it must contain points of arbitrarily large tastiness. Pick some lattice point $(c, d)$ in $B'$ with tastiness $\ell \ge 2 |a| + 10^{10^{10^{10}}}$. For $n \in \mathbb{N}$, define the $n-$ring to be the set of lattice points $(x, y)$ with tastiness $n$. Note that because $B$ is infinite, the $n-$ring always contains red points. For $|a| \le j \le \ell$, the $j-$ring has at least one blue point as well by considering a path from $(a, b)$ to $(c, d).$ Hence, for $|a| \le j \le \ell$, the $j-$ring has at least one empty square, because it has at least one red and at least one blue point, but no neighboring red and blue points. Thus, if we consider the $(4|a| + 2 \cdot 10^{10^{10^{10}}} + 5) \times (4|a| + 2 \cdot 10^{10^{10^{10}}} + 5)$ square centered at $(0, 0)$, we get an immediate contradiction. Hence, there's no other infinite branch, and the problem is solved. $\square$