Adding to what @pyramids conveyed in their answer:
A qubit's state is generally written as $\alpha|0\rangle + \beta|1\rangle$, where $\alpha, \beta \in \Bbb{C}$, and $|\alpha|^2+|\beta|^2=1$.
$\Bbb{C}^2(\Bbb{R})$ is a four-dimensional vector space, over the field of real numbers. Since any $n$-dimensional real vector space is isomorphic to $\Bbb{R}^n(\Bbb{R})$, you can represent any qubit's state as a point in a $4$-dimensional real space, too, whose basis vectors you can consider to be $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)$. In such a case a qubit's state would be represented as $a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1)$.
Say, $\alpha = a + i b$ (where $a,b\in \Bbb{R}$) and $\beta = c + id$ (where $c,d\in \Bbb{R}$). You need the condition $|a+ib|^2+|c+id|^2=1\implies a^2+b^2+c^2+d^2=1$ to be satisfied, which implies the state of the qubit would be a point on a 3-sphere.
As you know, it is difficult to efficiently represent a $4$-dimensional space on a $2$-dimensional surface like a paper, or your screen. Hence, you don't see that representation used often. Bloch sphere is pretty much the most efficient representation out there (for a single qubit), since it reduces one degree of freedom (of the complex numbers $\alpha,\beta$ each of which have two degrees of freedom) due to the fact that a qubit's state is usually normalized to a magnitude of $1$ i.e. $|\alpha|^2+|\beta|^2=1$.
Now, using the Hopf
coordinates
let's say:
$$\alpha = e^{i\psi}\cos(\theta/2)$$
$$\beta = e^{i(\psi+\phi)}\sin(\theta/2)$$
Here, $\theta$ can run from $0$ to $\pi$ whereas, $\psi$ and
$\phi+\psi$ can take values between $0$ to $\pi$.
In case you're wondering why $\theta/2$ is being used instead of $\theta$ have a look at the answers on this excellent thread on Physics Stack Exchange.
Okay, even now you notice three degrees of freedom $\psi,\phi,\theta$, whereas in a unit radii sphere, you only have two angles which you can change to get the different states of a qubit.
Notice that $\phi$ is basically the "relative phase" between $\alpha$ and $\beta$. On the other hand $\psi$ does not contribute to the "relative phase" of $\alpha,\beta$. Also, neither $\phi$ nor $\psi$ contribute to the magnitude of $\alpha,\beta$ (since $|e^{i\varphi}|=1$ for any angle $\varphi$). Since $\psi$ contributes neither to "relative phase" nor to the "magnitudes" of $\alpha,\beta$ it is said to have no physically observable consequences and we can arbitrarily choose $\alpha$ to be real by eliminating the factor of $e^{i\psi}$.
Thus we end up with:
$$\alpha = \cos(\theta/2)$$ and $$\beta=e^{i\phi}\sin(\theta/2)$$
Where $\theta$ can run from $0$ to $\pi$, and $\phi$ can run from $0$ to $2\pi$.
This practical simplification allows you to represent a qubit's state using just $2$ degrees of freedom on $3$-dimensional spherical surface having unit radius, which again can again efficiently be "drawn" on a $2$-dimensional surface, as shown in the following image.
Mathematically, it is not possible to reduce the degrees of freedom any further, and so, I'd say there is no other "more efficient" geometrical representation of a single qubit than the Bloch sphere.
Source: Wikipedia:Bloch_Sphere