Mathematically
The waves are solutions of the wave equation:
$$\Delta f - \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} = 0$$
This equation can be solved by many tools. The most elegant method is probably Fourier transformation; it allows us to separate the solution in coordinates (that's useful in some physics applications). The solution you mentioned $f = g(z-v/t) + h(z+v/t)$ is only one solution, but it doesn't cover the whole picture (understand the whole space of solutions).
Physically
We can find wave equation in few basics examples. The first one could be the string vibrating for small initial deviations. The other wave equation can be found in Maxwell equation for field $\vec E, \vec B$ or for scalar and vector potentials $\phi, \vec A$. It only means that these waves are physical, but these waves must still satisfy Maxwell equations.
In physics, this is only a special case of what we call waves. We have Klein-Gordon waves, Dirac waves (you will learn this in quantum field theory) or very simple equation from electrodynamics: waves in conductors. All these waves don't satisfy the wave equation in the classical meaning of \eqref{A}. But we can still call it waves.
So mathematically waves are strictly solutions of the wave equation \eqref{A}, physically we call disturbance in the space that are time-dependent or physical entities carrying information that is propagating from one place to another.
The word "wave" has its origin. Mathematicians in history (in post-renessaince era mostly) were working on the description of the musical instruments. So the first origin of waves was from mathematicians studying physical nature. I can recommend you this article [*].
Edit:
When Maxwell equations are reduced into wave equations, it really means that electromagnetic waves exists. That is because of existence of $\vec E, \vec B$. These fields exists, they carry momentum and energy. So if they are solution of the wave equation, they must be waves.
This led many of physicists to think that there is an aether, which is a medium where electromagnetic waves (as light) can be disturbance. And this is the problem known in special relativity when Michelson and Moorley proved that aether (even if it exists) is not important for electromagnetic waves. This experiment was proof for Einstein special relativity and it started the modern era of the physics.
[*] https://www.jstor.org/stable/41134001?seq=1#page_scan_tab_contents
Best Answer
Mason handled the distinction between inhomogeneous and homogeneous differential equations, but if one is speaking of the most general possible form of the wave equation, it is,
$$\square\phi^{i_1\dots i_m}_{j_1 \dots j_n}(x) = f^{i_1 \dots i_m}_{j_1 \dots j_n}(x)$$
where both fields are rank $(m,n)$ tensors, acted upon by the Laplace-Beltrami operator $\square = \nabla^a \nabla_a$ whose action on the tensors depends on both the metric and their rank. For a scalar field with metric $\eta_{\mu \nu}$, it reduces to the most familiar form of the wave equation, $(\partial^2_t - \nabla^2)\phi = f$. (The above can also be recast in the language of differential forms.)
However, in a way this does not cover all the possibilities. For example, in general relativity, for a perturbation $h_{ab}$ of the metric, the first order change in the curvature is,
$$\delta R_{ab} \propto \Delta_L h_{ab} = \square h_{ab} -2 \nabla_{(a} \nabla^c \bar{h}_{b)c} -2 R_{d(a} h^d_{b)} +2 R_{acbd}h^{cd}$$
which is understood as the curved space 'wave operator' in the literature because it certainly admits wave solutions but is clearly not equivalent to the wave equation above as it contains other terms involving curvature tensors. Thus, the 'most general form' of the wave equation isn't something we can really write down, unless your idea of it is strictly $(\partial^2_t - \nabla^2)\phi = f$.