(*This is a partial answer. Courtesy of a comment by Nemo.*) 
>**Part I.** $x^5+y^5+z^5 = 1$
 
It turns out that just like,
$$x^3+y^3=1$$
can be solved by the cubic continued fraction $C(q)$ and $C(q^3)$, its quintic analogue
$$x^5+y^5+z^5 = 1$$
has a beautiful solution using the Rogers-Ramanujan cfrac $R(q)$ and $R(q^5)$. Given,
$$R(q)=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$

then we have the simple,

$$\alpha^5\phi^5+\beta^5\phi^5+\alpha^5\beta^5 = 1\tag1$$

where,

$$\alpha = R(q),\quad\beta = \frac{1-\phi R(q^5)}{\phi+R(q^5)}$$
and ***golden ratio*** $\phi$. This was derived by yours truly using Tim Huber's "[*A Theory of Theta Functions to the Quintic base*][1]". 
>**Part II.** $x^5+y^5 = 1$

Huber  defines four functions which can be ultimately expressed in terms of the [Rogers-Ramanujan identities][2]. Define $q=e^{2\pi i z}$ and,
$$P(z):=q^{11/60}H(q)=q^{11/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}$$
$$Q(z):=q^{-1/60}G(q)=q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}$$
then Huber's four functions in simplified form are,
$$\begin{aligned}
a(\tau) &=\eta^{2/5}(\tau)\;P(\tau)\\[2mm]
b(\tau) &=\eta^{2/5}(\tau)\;Q(\tau)\\[2mm]
c(\tau) &= 5^{1/4}\phi^{1/2}\,\eta^{2/5}(5\tau)\;P\big(\tfrac{-1}{5\tau}\big)\\[2mm]
d(\tau) &= \frac{5^{1/4}}{\phi^{1/2}}\,\eta^{2/5}(5\tau)\;Q\big(\tfrac{-1}{5\tau}\big)\end{aligned}$$
which obeys,
$$\Big(\frac{a\,\phi}{b}\Big)^5+\Big(\frac{c}{b}\Big)^5 = 1\tag2$$
$$-\Big(\frac{a}{b\,\phi}\Big)^5+\Big(\frac{d}{b}\Big)^5 = 1\tag3$$
It then follows that the ratio of $a,b$ is the Rogers-Ramanujan cfrac,
$$\frac{a(\tau)}{b(\tau)} = R(q)$$
and $R(q)$ ***in a way*** can parameterize the Fermat quintic. Some manipulation will also show that,
$$\frac{c(\tau)}{d(\tau)} = \phi\,\frac{1-\phi\, R(q^5)}{\phi+R(q^5)}$$
Thus, combining $(2),(3)$ will yield $(1)$.

  [1]: https://arxiv.org/pdf/1304.0684.pdf
  [2]: https://en.wikipedia.org/wiki/Rogers%E2%80%93Ramanujan_continued_fraction