The hydrogen atom
Indeed, three numbers cannot define a function. There are several more criteria, one of which you mentioned already: it must satisfy the time-independent Schrödinger equation. The Schrödinger equation is a nasty differential equation, which looks something like the following:
$$-\frac{\hbar^2}{2\mu}\left(\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}\right) - \frac{e^2}{4\pi\varepsilon_0r}\psi = E\psi$$
On top of this, it must also satisfy the usual criteria for a wavefunction ("boundary conditions"): for example it must be square integrable (i.e. its integral over all space must be finite, meaning that it can be normalised):
$$\int |\psi|^2\,\mathrm{d}\tau < \infty$$
Once you put all of these conditions together, it turns out that not every plain old function $\psi(x,y,z)$ is a permissible atomic orbital. There is a family of functions $\psi$ which do, and they all have a certain form which is completely determined by the three quantum numbers $(n, l, m_l)$.
A simpler example
This is pretty abstract, so why don't we make it more concrete with a simpler example. (The hydrogen atom is not easy to solve explicitly.) Consider the one-dimensional function $f = f(\phi)$, where $\phi$ is an angle. You should recognise at this point that, given no other information, one quantum number is not sufficient to specify what $f$ is, as there is an infinite possible different $f$'s.*
However, assume now that $f$ obeys the differential equation
$$\frac{\mathrm{d}f}{\mathrm{d}\phi} = \mathrm{i}mf$$
The general solution of this differential equation is (after dropping the normalisation factor)
$$f = \exp(\mathrm{i}m\phi)$$
But note also that since $\phi$ is an angle, we have that for all $\phi$
$$f(\phi) = f(\phi + 2\pi)$$
meaning that
$$\begin{align}
\exp(\mathrm{i}m\phi) &= \exp[\mathrm{i}m(\phi + 2\pi)] \\
&= \exp(\mathrm{i}m\phi) \exp(\mathrm{i}m2\pi) \\
1 &= \exp(\mathrm{i}m2\pi)
\end{align}$$
which means that $m$ must be an integer. Consequently, the allowed $f$'s are
$$f = \exp(\mathrm{i}m\phi) \quad\quad m \in \mathbb{Z}$$
and so we can now label the $f$'s by the value of $m$. The quantum numbers $m$ do not represent actual values of $f$ at any point in space, nor do they represent actual values of $\phi$: they instead serve to parameterise the $f$'s. That is to say, the value of $f$ at any value of $\phi$ is completely determined by the value of $m$. We do not write this as $f(m)$, but rather $f_m$, and thus we have:
$$\begin{align}
&\,\,\,\vdots \\
f_{-2} &= \exp(-2\mathrm{i}\phi) \\
f_{-1} &= \exp(-\mathrm{i}\phi) \\
f_{0} &= 1 \\
f_{1} &= \exp(\mathrm{i}\phi) \\
f_{2} &= \exp(2\mathrm{i}\phi) \\
&\,\,\,\vdots
\end{align}$$
Indeed, $m$ can be said to be a quantum number here. Notice how we've used a simplified problem: it has one instead of three dimensions, and it has a much simpler differential equation. However, it illustrates that by enforcing additional conditions on the wavefunction — namely that it obey a differential equation and boundary conditions — we can get a family of solutions which can be labelled with quantum numbers.†
What is the energy?
When you find the allowed $\psi$'s that solve the Schrödinger equation, you can plug it back into the differential equation and evaluate the value of $E$ inside it. That will give you the energy of the atomic orbital.
Footnotes
* Technically, there is an uncountably infinite number of $f$'s. If it were countably infinite, then we could establish a one-to-one mapping between the different $f$'s and the natural numbers, which would allow us to label the $f$'s as $f_1, f_2, f_3, \cdots$ which is exactly what a quantum number does.
† With regards to the previous footnote, this set of permissible $f$'s is still infinite, but it is countably infinite this time. Also, this actually happens to be the exact way in which the quantum number $m_l$ is derived, so my choice of $m$ as the quantum number was not arbitrary.
