On solvable octic trinomials like $x^8-5x-5=0$ Solvable quintic trinomials
$$x^5+ax+b=0$$
have been completely parameterized. Finding $6$th-deg versions is relatively easy to do such as,
$$x^6+3x+3=0$$
which factors over $\sqrt{-3}$. No $7$th-deg are known, but surprisingly there are octic ones, such as the simple,
$$x^8-5x-5=0$$
which factors over $\sqrt{5}$. And the not-so-simple ones,
$$x^8-11(4x+3)=0\\x^8+16(4x+7)=0\\x^8 + 5\cdot23^2(12 x+43) =0$$
which factors over a quartic extension (and needs the cube root of unity).

Q: Any other octic examples, if possible parametric?


$\color{green}{Update:}$
Klajok in his answer below has found a family for the class of octic trinomials that factor over a quadratic extension. However, another class needs a quartic extension. For example,
$$x^8-44x-33=0\tag1$$
which factors into four quadratics,
$$x^2 + v x - (2v^3 - 7v^2 + 5v + 33)/13=0$$
and where $v$ is any root of $v^4 + 22v + 22=0$. More generally, eliminating $v$ between
$$x^2 + v x + (pv^3 +qv^2 + rv + s)=0$$
$$v^4+av^2+bv+c=0$$
easily done by the resultant function of Mathematica will result in an irreducible but solvable octic and judicious choice of rational coefficients will yield a trinomial. However, it is not known if this second class of trinomials like $(1)$ has a parametric family as well.

A result of Harris [1] is that every monic palindromic polynomial of degree-8 can be factored into two monic palindromic polynomials of degree-4.
$$
\begin{align}
f(x) & = x^8 + ax^7 + bx^6 + cx^5 + dx^4 + cx^3 + bx^2 + ax + 1 \\
     & = (x^4 + px^3 + qx^2 + px + 1)(x^4 + rx^3 + sx^2 + rx + 1) \\
     & = x^8+x^7 (p+r)+x^6 (pr+q+s)+x^5 (p(s+1)+qr+r)+x^4 (2pr+qs+2)+x^3 (p(s+1)+qr+r)+ x^2 (pr+q+s)+x (p+r)+1
\end{align}
$$
Equating the coefficients:
$$
\begin{align}
a &= p+r \\
b &= pr + q + s \\
c &= p(s+1) + qr + r \\
d &= 2pr + qs + 2
\end{align}
$$
For the subset of degree-8 monic palindromic polynomials of the form
$$
\begin{align}
f(x) & = x^8 + 0x^7 + 0x^6 + 0x^5 + dx^4 + 0x^3 + 0x^2 + 0x + 1 \\
     & = x^8 + dx^4 + 1 \\
\text{we have }
0 &= p+r \\
0 &= pr + q + s \\
0 &= p(s+1) + qr + r \\
d &= 2pr + qs + 2
\end{align}
$$
We have the parametric solutions:
$$
r = 0 ∧ s = -q ∧ p = 0 ∧ d = 2 - q^2, \\
r = \sqrt{2} \sqrt{q} ∧ s = q ∧ p = -\sqrt{2} \sqrt{q} ∧ d = q^2 - 4 q + 2, \\
r = -\sqrt{2} \sqrt{q} ∧ s = q ∧ p = \sqrt{2} \sqrt{q} ∧ d = q^2 - 4 q + 2. \\
$$
Note that all of the above are parametrized on $d$.
On similar lines, we can factor the monic quartic palindromic polynomials into two monic quadratic palindromic polynomials each and then use the quadratic formula to get the roots.
References
[1]: J. R. Harris, "96.31 Palindromic Polynomials," The Mathematical Gazette, vol. 96, no. 536, p. 266–69, 2012. https://doi.org/10.1017/S0025557200004526
