Multiplying integers modulo $2^{255}-19$ using the Curve25519 polynomial reduction algorithm
The paper states


  $R$ contains $2^{255}x^{10}-19$, which represents $0$ in $\mathbb Z/(2^{255}-19)$.


In other (slightly simplified) words:
  $$x^{10} \;\;\text{is}\;\; 2^{-255}\cdot19 \text!$$
(Note that this contradicts your comment to the question: $x^{10}$ does not represent $2^{255}$.)

To understand this, recall that a value in $\mathbb Z/(2^{255}-19)$ is represented by a polynomial which has that value at the point $1$. Indeed, evaluating $2^{255}x^{10}-19$ at $1$ yields precisely the modulus $2^{255}-19$, hence identifying it with zero does not change the represented value.
Now observe that reducing modulo $2^{255}x^{10}-19$ means identifying that polynomial with zero.
The resulting relation can also be stated as
$$\begin{align*}
&amp;2^{255}x^{10}-19 \;=\; 0
\\\iff\quad&amp; 2^{255}x^{10} \;=\; 19
\\\iff\quad&amp; x^{10} \;=\; 2^{-255}\cdot 19
\text, \end{align*}$$
as claimed. Thus you can always (for instance: when reducing degrees) replace $x^{10}$ by $2^{-255}\cdot19$ without changing the represented value.