When does iterating $z \mapsto z^2 + c$ have an exact solution?
No, there are no others.
Analytically, one can show that if a Julia set contains an analytic arc, it is in fact a straight line or a circle (up to conjugation). For the class $z^2+c$, $0$ and $-2$ are the only values where this occurs.
This does not quite imply that there are no closed-form solutions of the recurrence for any other value of $c$, but such a closed form would naturally describe a fractal. As far as I know, there are no "closed form" functions which directly have a fractal as one of its level set.
I'd like to add a little more context for the record here, to amplify and extend Jacques Carrette's answer as well as the extra information in the comments:
The Chebyshev polynomials, together with their close relatives the maps $z \to z^n$, are the the only polynomials exhibiting the property described in the question, up to affine coordinate change. Geometrically, the map $z \to z^n$ is an $n$-fold covering map of the cylinder $\mathbb C \setminus 0$ over itself. This cylinder has an order 2 symmetry $z \to z^{-1}$ that commutes with the covering map. The quotient of the cylinder by the symmetry is equivalent to $\mathbb C$, or preserving a little more information, it's an orbifold structure on $\mathbb C$, with two order 2 cone points at $\pm 2$.
The expression $z + z^{-1}$ is the formula for the quotienting map $\mathbb C \setminus 0 \to \mathbb C$. The cylinder can be unwrapped further to the universal cover (also $\mathbb C$) of $\mathbb C \setminus 0$; the quotient maps from the universal cover to the two quotients are $\exp$ and $\cos$, and so the Chebyshev polynomials (up to affine normalization) are the formulas for $\cos$ of a multiple angle in terms of $\cos$ of an angle.
There are however other rational functions which admit explicit analytic formulas for iteration. These are called Lattès examples, named after Samuel Lattès who discovered them in 1918 (although according to Milnor's article in the Bodil Branner festschrift, Schöder described similar examples almost 50 yeasr earlier, but they appear to have been forgotten).
The Lattès examples are symmetry quotients of isomorphisms of wall-paper groups to proper subgroups. For every orientation-preserving wall-paper group except the group of translations, the quotient space (wallpaper modulo symmetry) is topologically a sphere. Every such group is isomorphic to subgroups of itself in infinitely many ways. These isomorphisms descend to branched coverings of the quotient space over itself: in the language of complex functions these are rational maps, in the language of orbifolds these are self-coverings.
They have a significant role as exceptional examples in complex dynamics, and are well understaood, with lots of nice pictures available someplace in the literature. For example, an endomorphism of the tiling groups whose formula is multiplication by the complex number $\lambda$ can be used for a kind of base $\lambda$ expansion of complex numbers, associated with a self-similar tiling of $\mathbb C$ (with basic tile is those numbers starting after the lambdimal point). At least for some cases, the digits can be arranged in an order that defines a sphere-filling curve invariant under multiplication by $\lambda$.
All these examples have explicit analytic formulas for iterations, completely analogous to the formula $z \to 2 \cos (k^n \arccos ( z/2))$ for the $n$th iterate of the $k$th Chebyshev polynomial. In the Chebyshev formula, $\cos$ is the universal covering map of $\mathbb C$ for the $(22\infty)$ orbifold. For wallpaper groups, the universal covering maps are elliptic functions instead of $\cos$. The rational map lifts to multiplication by a complex number $\lambda$, so you lift to the cover, multiply by a power of $\lambda$, and map back down.
Actually, there are many other cases : if $f$ is a surjective map from $C$ to $C$ such that $f(az)=f(z)^2+c$, we get the simple formula $z_n=f(a^nZ0)$, where $f(Z_0)=z_0$. So the question is : for which values of $a$ and $c$ does such an $f$ exists, with perhaps some more constraints on $f$ such as smoothness. It turns out that for any $c$ large enough (at least of module greater than 1) there exists exactly one such entire function $f$, with $f'(0)=1$ and $a=2f(0)=1+\sqrt{1-4c}$ (this is shown by an elementary, but tedious calculation of the coefficients of the series by recurrence (done in French here, page 8)), followed by a simple argument of domination of those coefficients, proving convergence for $R>0$ ; then, analytic continuation shows that the radius of convergence is infinite) ; $f$ is (almost) surjective by Picard theorem. Of course, those $f$ are never "usual" or elementary functions, except for $c=0$ or $c=-2$, but so what?