I'm going to rewrite this answer to clarify what I think the issue is. I think the OP is imagining a different definition of the ring $k[x]$ than most answerers are. Here are two reasonable definitions:
$k[x]$ is the ring of formal expressions of the form $\sum_{j=0}^{\infty} p_j x^j$ with $p_j \in k$ and we require that $p_j$ is $0$ for $j$ sufficiently large. See Wikipedia for the details.
Let's define $\mathrm{Poly}(k)$ to be the ring of functions $k \to k$ which are of the form $x \mapsto \sum_{j=0}^d p_j x^j$ for some $p_j \in k$.
In any book on modern algebra which is sophisticated enough to have careful definitions (Herstein, Artin, Dummit and Foote...), the definition of $k[x]$ will be something like the first one. With this definition, it is very clear that degree is well defined because equality of polynomials is defined to mean equality of coefficients.
But the OP talking about a polynomial having two different representations makes me think that he or she is imagining a definition where that might be possible, and Definition 2 strikes me as the most likely. Indeed, if $k$ is the field with $q$ elements, then $x^q$ and $x$ represent the same function in $\mathrm{Poly}(k)$, even though one is a polynomial of degree $q$ and the other has degree $1$. This is one of the reasons that Definition 1 is the standard definition.
In a course which is sophisticated enough to have careful definitions, it would be appropriate to prove that the obvious map $k[x] \to \mathrm{Poly}(k)$ is an isomorphism for $k$ infinite. I could also imagine a course where the ground field is always $\mathbb{R}$ and one might prove that $\mathbb{R}[x] \cong \mathrm{Poly}(\mathbb{R})$ without bringing up that finite fields exist.
But we don't have to raise the issue when we define degree, because Definition 1 is the standard definition of $k[x]$ and this avoids the issue. Also, as Henry Townser says, many courses are at too low a level to reach this level of precision.