but it's orders of magnitudes more intuitive [...]
I would dispute this point, personally. Imagining a tiny ball guided by a pilot wave is perhaps easier to visualize than a non-localized quantum state, but visualizing a picture in your head does not equate to an intuitive understanding of the theory, especially when the dynamics are governed by something as weird as a pilot wave - this may become more clear momentarily.
The full answer to your question goes as follows (skip to the end for the summary):
The pilot wave $\psi:\mathbb R^3\rightarrow \mathbb C$ can be expressed as $\psi(\mathbf x) = R(\mathbf x) e^{iS(\mathbf x)/\hbar}$ for $\mathbb R$-valued functions $R$ and $S$.
In de Broglie-Bohm theory, a particle is subject not only to an external classical potential $V$ but also "quantum potential" $U$ defined by
$$U(\mathbf x) = -\frac{\hbar^2}{2m}\frac{\nabla^2R(\mathbf x)}{R(\mathbf x)}$$
The momentum of a particle at a point $\mathbf x$ is uniquely determined by the pilot wave and is given by
$$\mathbf p(\mathbf x) = \nabla S(\mathbf x)$$
Therefore, the energy of a particle at $\mathbf x$ is given by
$$E(\mathbf x) = \frac{\mathbf p^2}{2m} + V(\mathbf x) + U(\mathbf x) = \frac{[\nabla S(\mathbf x)]^2}{2m} + V(\mathbf x) - \frac{\hbar^2}{2m} \frac{\nabla^2 R(\mathbf x)}{R(\mathbf x)}$$
Note that
$$\nabla^2\psi = \nabla \cdot(\nabla \psi) = \nabla \cdot \left([\nabla R]e^{iS/\hbar} + \frac{i[\nabla S]}{\hbar} Re^{iS/\hbar}\right)$$
$$= [\nabla^2 R]e^{iS/\hbar} + \frac{i\nabla S\cdot \nabla R}{\hbar} Re^{iS/\hbar} - \frac{[\nabla S]^2}{\hbar^2}Re^{iS/\hbar} + \frac{i\nabla^2 S}{\hbar} Re^{iS/\hbar}$$
Requiring that $R\rightarrow 0$ as $\mathbf x\rightarrow \infty$ and integrating by parts, we can cancel the second term against the fourth and obtain
$$\frac{\nabla^2\psi}{\psi}=\frac{\nabla^2 R}{R}-\frac{[\nabla S]^2}{\hbar^2}$$
Therefore, we see immediately that
$$E(\mathbf x) = \frac{1}{\psi(\mathbf x)}\left[-\frac{\hbar^2}{2m}\nabla^2 \psi(\mathbf x) + V(\mathbf x)\psi(\mathbf x)\right]$$
The states of fixed energy are the ones such that $E(\mathbf x) = E_0$ is constant at every point in space - which implies that the pilot wave satisfies
$$-\frac{\hbar^2}{2m} \nabla^2 \psi(\mathbf x) + V(\mathbf x)\psi(\mathbf x) = E_0 \psi(\mathbf x)$$
Solving this equation then yields the possible states of constant energy.
Obviously my answer is not necessarily intended for a high-school audience. It can be summarized less technically as follows:
In de Broglie-Bohm theory, the particle is subject to a "quantum potential" in addition to the classical Coulomb potential you'd normally include when studying the hydrogen atom. Both this quantum potential and the particle's momentum at any particular point are determined by the pilot wave $\psi$.
When we add up the kinetic energy, classical potential energy, and quantum potential energy of the particle at a point, we find that the total energy is given by $E(\mathbf x)=\big[-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf x) + V(\mathbf x)\psi(\mathbf x) \big]/\psi(\mathbf x)$.
If you want to look for states of fixed energy, then you solve
$$E(\mathbf x)=E_0 \iff -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf x) + V(\mathbf x)\psi(\mathbf x) = E_0 \psi(\mathbf x)$$
which is precisely the time-independent Schrodinger equation from standard quantum mechanics.
To distill it down even more simply, de Broglie-Bohm theory is exactly the same as standard quantum mechanics from an operational and mathematical standpoint. The only difference is in the interpretation. In both cases, states of definite energy are obtained by solving the equation
$$-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf x) + V(\mathbf x)\psi(\mathbf x) = E_0 \psi(\mathbf x)$$
In standard QM, $\psi$ is the wavefunction of the particle; in de Broglie-Bohm, it is the pilot wave, which determines the particle's momentum and applies a "quantum potential" which are both engineered specifically to produce the same results as standard QM.
