The main paper you want to consult is
É. Cartan, Familles de surfaces isoparamétriques dans les espaces à courbure constante, Annali di Mat. 17 (1938), 177–191.
In this paper, Cartan considers the problem of studying the functions $f$ defined on an open set in a space $M$ of constant curvature that satisfy two equations of the form $|\nabla f|^2 = A(f)$ and $\Delta f = B(f)$ for some functions, $A>0$ and $B$, of one variable. He shows that the (hypersurface) level sets of $f$ are what he called isoparametric, i.e., they have all of their principal curvatures constant.
This had been considered somewhat earlier earlier by Levi-Civita for the case $n=3$ in flat space and by Segre for all $n$ (again, in flat space). They showed that, up to rigid motion and homothety, a connected isoparametric hypersurface must be an open subset of $S^k\times \mathbb{R}^{n-k-1}$ for some $0\le k\le n{-}1$.
Cartan showed that, when the ambient sectional curvature is negative, i.e., in hyperbolic $n$-space, then, again, a connected isoparametric hypersurface must either be totally umbilic (i.e., an open subset of either a totally geodesic hyperplane, an equidistant hypersurface, a horosphere, or a hypersphere) or be an open subset of a hypersurface of the form $S^k\times H^{n-k-1}$, i.e., it must be a totally geodesic hypersurface or horosphere or else the set of points of constant distance from a totally geodesic submanifold of lower dimension. Thus, if you did have a solution of $|\nabla f|^2=1$ and $\Delta f = 0$ (even a local one in a neighborhood of $p$) in hyperbolic $n$-space, then $f-f(p)$ would be the oriented distance from the level set $f=f(p)$, and it is easy to calculate that no such function is harmonic, so you have a contradiction.
By the way, there is a much greater variety of isoparametric hypersurfaces in spaces of constant positive curvature. In fact, even though Cartan found many nontrivial examples, and there has been a lot of work since then constructing more examples and classifying them, the classification of these is still not complete.