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If $R$ is a $\mathbb{Z}$-graded ring with unit that is a finite-dimensional vector space over some field $\mathbb{K}$ (for example $\mathbb{Q}$), then is it true that all elements of $R$ have degree zero?

I tried to make the following argument but don't know if it is correct. Suppose $a_1, \cdots, a_n$ are generators of $R$ over $\mathbb{K}$ and are homogeneous with respect to the grading (not sure whether this is always possible!) and for all $i$, $-C < |a_i| < C$ for some integer $C$. Let $x \in R$ with $|x| \ne 0$. Then $|x^m| = m|x|$ goes to infinity as $m$ goes to infinity. But also $x^m = k_{1, m}a_1 + \cdots + k_{n, m}a_n$ for some $k_{i,m} \in \mathbb{K}$. Since $a_i$ have degree bounded by $\pm C$, the same must be true for $x^m$, which is a contradiction. So $|x| = 0$.

Furthermore, is it possible to have graded, unital ring maps $\phi: R_1 \rightarrow R_2$, where $R_1$ is finite-dimensional but $R_2$ is infinite-dimensional over $\mathbb{K}$?

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    The ring k[X]/(X^2) can be graded so that (the class of) X is in degree 1. – 2017-02-20
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    @MarianoSuárez-Álvarez Ok, maybe the right question is whether all elements either are nilpotent or are in degree zero? I'll make an edit. – 2017-02-20
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    As for the scond question, sure. Consider the ring $k[x,y]/(x^2)$, graded so that $x$ and $y$ have degree $1$. There is an injection fromby previous example to this one. – 2017-02-20
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    @MarianoSuárez-Álvarez, thank you! Also, I realized that if if $F$ is any unital ring over a field $K$, then we have a ring map $K \rightarrow F$; $K$ is finite-dimensional (dimension one) over $K$ and $F$ can be taken to be infinite-dimensional over $K$ . – 2017-02-20

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What's true is that all homogeneous elements of nonzero degree are nilpotent. Your argument implicitly assumes that $x^m$ is always nonzero.