How exactly do quantum numbers increase in relation to energy levels in more than one dimension?

For a particle in a box (also known as a particle in an infinite potential well) in $d$ dimensions, the Hamiltonian inside the box is given by $$hat{H}=frac{1}{2m}sum_{i=1}^d{p}^2_i$$where $p_i$ is the momentum operator in the $i^{text{th}}$ direction. As you already know (I infer so from your question) that using the boundary conditions, we arrive at the conclusion that $p_i$ must be of the form $$p_i=frac{n_ipi}{L_i}$$where $L_i$ is the length of the box in the $i^{text{th}}$ direction and $n_iinmathbb{N}$. Thus, the energy of an eigenstate is given by $$E=frac{pi^2}{2m}sum_{i=1}^dfrac{n_i^2}{L_i^2}$$.

Now, in the case of one--dimensional box, $E$ only depends on $n_x$ and is monotonically increasing with $n_x$. This means that the spectrum of a one--dimensional particle in a box is non--degenerate, and the value of the energy depends only on one quantum number, namely, $n_{x}$. One can simply write that $E=E_{n_x}$

However, in the case of a multidimensional box, $E$ depends on the multiple quantum numbers, namely, ${n_i}$ where $i>1$. And, we have to write $E=E_{{n_i}}$ to signify that $E$ depends on multiple quantum numbers.

Moreoever, multiple different sets of values of $n_i$ can produce the same value of energy. For example, let's consider the case where $L_i=L$ and $i=3$. All the following configurations of $n_i$ produce the same value of energy:$$n_x=2,n_y=1,n_z=1$$ $$n_x=1,n_y=2,n_z=1$$ $$n_x=1,n_y=1,n_z=2$$ Thus, we see that the spectrum of a particle in a box can be degenerate in a higher dimension.

So, to answer your question, the first excited energy state in a three-dimensional particle in a box (with equal sides) can correspond to all of the above configurations of quantum numbers.