If a group is q.i. to $\mathbb Z$, then all its asymptotic cones are lines. In our paper with Cornelia Drutu, "Tree-graded spaces and asymptotic cones of groups", Topology 44 (2005), no. 5, 959--1058 (see Corollary 6.2), we prove that if even one asymptotic cone of a group is a line, then the group is virtually cyclic.

Now about the original question. We need to show that if the growth of $G$ is linear, say ~$Cn$, then an asymptotic cone of $G$ is a line. Since the group is infinite, it a sequence of arbitrary long geodesics $g_n$ of length $n$ with midpoint $1$. Now pick any number $m$ and two points $a,b$ at distance $2m$ on $g_n$, $n\gg m$, let $c$ be the midpoint between $a,b$ on $g_0$. The balls $B(a,m)$, $B(b,m)$ then contain $~Cm$ points each, hence their (disjoint!) union is (very close to) the ball $B(c,2m)$. Hence the $2m$-neighborhood of $g_n$ is contained in the $m$-neighborhood of $g_0$. This should be true for a sequence of $m_i$'s tending to $\infty$. From this it is easy to deduce that the asymptotic cone ${\mathcal C}=Con^\omega(G,(m_i))$ is at distance $1$ from the geodesic line $g$ which is the limit of $g_n$. But then ${\mathcal C}$ is quasi-isometric to a line, hence its asymptotic cone is a line. It is proved in the paper cited above that an asymptotic cone of $\mathcal C$ is again an asymptotic cone of $G$, and we are done.