Let $\Gamma$ be a tree, and $m \ge 2$. If $\Gamma$ is an $m$-tree (all vertices have valence of $m$), then $\Gamma$ is infinite.
My idea is to prove it by contradiction. Suppose that $\Gamma$ is finite yet every vertex has a valence of $m$. I suspect that I should be able to contradict the formula $|V(\Gamma)| = |E(\Gamma)| + 1$, but I don't know how to count $|E(\Gamma)|$. Given $n := |V(\Gamma)|$, what is $|E(\Gamma)|$? I'm having hard time determining this. I tried several examples, but I couldn't discern a pattern/formula.
It doesn't have to be that complicated. You can directly construct an infinite path in $\Gamma$. Start at some vertex $v_0$. Since $v_0$ has valence $m \geq 2$, we can choose a neighbor $v_1$ of $v_0$. Now $v_1$ has valence $m \geq 2$, so we can choose a neighbor $v_2 \neq v_0$ of $v_1$. Now $v_2$ has valence $m \geq 2$, so we can choose a neighbor $v_3 \neq v_1$ of $v_2$. We have to check that $v_2 \neq v_0$ though... why can't this happen? Continue in this fashion.