In abstract algebra, we write the set of all polynomials with coefficients in a ring $R$ as $R[x]$.
Here "polynomials" means expressions of the form $$a_0+a_1x+a_2x^2+\cdots +a_nx^n$$ where $a_0,\ldots,a_n\in R$ and $n$ is finite (Note: if you don't know what a ring is, just think of the $a$'s as numbers). So, in this context, your expression $x^{-4}+x^3$ isn't in a polynomial, because all the powers of $x$ have to be non-negative.
We can generalize this construction, though. The first thing we can do is drop the requirement that $n$ must be finite. If we do this, we get $R[[x]]$, the set of formal power series in $R$.
A futher generalization is the set of formal Laurent series in $R$, denoted $R(\hspace{-0.5pt}(x)\hspace{-0.5pt})$, and this is a setting in which we can answer your question. Formal Laurent series have the form
$$\sum_{n\in \mathbb{Z}}a_nx^n$$
where $a_n=0$ for all but finitely many negative $n$. In other words, formal Laurent series are formal power series which are allowed to have a finite number of negative exponents too.
The order of a formal Laurent series is defined as the smallest $n$ such that $a_n\not= 0$. This is kind of like the degree of a polynomial, but for negative integers. The degree of a formal Laurent series is defined in the same way as the degree of a polynomial, though the degree may not exist (since all of the $a_n$ for $n>0$ are still allowed to be nonzero).
So, considered as a formal Laurent series, we would say that $x^{-4}+x^3$ has degree $3$ and order $-4$.
You have put your finger precisely on the statement that is incorrect.
There are two competing conventions with regard to rational exponents.
The first convention is to define the symbol $a^x$ for $a > 0$ only. The symbol $\sqrt[n]{a}$ is defined for negative values of $a$ so long as $n$ is odd, but according to this convention, one wouldn't write $a^{1/n}$, for instance.
In defining $a^{p/q}$ to be $(\sqrt[q]{a})^p$, the author you quoted chose the fraction $p/q$ to be in lowest form so that the definition would be unambiguous. For example, $a^{10/15}$ is defined to be $(\sqrt[3]{a})^2$. However, it is preferable to define $a^{p/q}$ to be $(\sqrt[q]{a})^p$ in all cases and to prove that this definition is independent of the particular representation chosen for $p/q$; this is what more rigorous books tend to do. That is, you prove that if $p/q = r/s$, then $(\sqrt[q]{a})^p = (\sqrt[s]{a})^r$. There is no mention of lowest form.
The competing convention is to also allow $a^x$ to be defined for all $a \ne 0$ and all rational numbers $x = p/q$ that have at least one representation with an odd denominator. You then prove that $(\sqrt[q]{a})^p$ is independent of the particular representation $p/q$ chosen, so long as the denominator is odd. Thus you can write $a^{3/5} = (\sqrt[5]{a})^3 = (\sqrt[15]{a})^{9} = a^{9/15}$. All of that is fine. However, you cannot write $a^{6/10} = (\sqrt[10]{a})^6$, or even $a^{6/10} = \sqrt[10]{a^6}$. The number $a^{6/10}$ is well-defined, but to write down its definition, you must first select a fraction equivalent to $6/10$ that has an odd denominator, which could be $3/5$ or $9/15$ or something else. For $a^{1/2}$, this can't be done at all, so $a^{1/2}$ is undefined for $a < 0$.
The rules for exponents break down if you start allowing $a < 0$ and exponents that can't be written with an odd denominator. For example, the rule $a^{xy} = (a^x)^y$ is valid, but only so long as $x$ and $y$ are both rational numbers that can be written with an odd denominator. This is not the case if you write $a^1 = (a^2)^{1/2}$, despite the fact that both sides of the equation are defined since $a^2 > 0$.
Edit Reading the paper by Tirosh and Even, I was surprised to learn this matter has drawn serious attention from math educators.
A long time ago, I assumed that, apart from complex extensions, $a^x$ for non-integer $x$ should be defined only for $a > 0$. I reasoned that it made no sense to have a function $(-2)^x$ defined only for rational numbers $x$ with odd denominator. I objected strenuously to notations like $(-8)^{1/3}$.
But that was before I taught a calculus class, which is when I realized why some textbook authors are so happy to define $a^x$ for $a < 0$, following the second convention. The reason is that the formula $\frac{d}{dx}(x^r) = rx^{r-1}$ is perfectly valid for $x < 0$ and $r$ with odd denominator.
Best Answer
One definition that is sometimes used is that the degree of a function $f(x)$ is given by
$$\text{degree}(f) = \lim_{x \to \infty} \frac{\ln | f(x) |}{\ln(x)}$$
For $x^r$, this nicely evaluates to $r$, even for non-integer $r$. You can also use this to see that, for $f$ that is the sum of such terms, whichever $r$ is the highest is the degree.
You can also use this for other functions, too, and find some degrees even if the function is not a polynomial in any realistic sense. For instance:
A bit more can be read on Wikipedia here.