Could the metric be a tensor product of a 4-vector with itself? I am really keen on the idea of geometrizing all of physics using the spacetime manifold, so I was happy to find this paper:
https://doi.org/10.1088/1742-6596/1956/1/012017
As I understand it, they demonstrate that Maxwell's equations are a consequence of the vacuum Einstein equation, IF you suppose that the following relation between the metric and the electromagnetic 4-potential:
$g_{\mu\nu} = A_{\mu}A_{\nu}$
They further identify 4-current with $\nabla_{\mu}\nabla^{\nu}A_{\nu}$ and explain how this relates to Weyl curvature.
Now I am eager to subscribe to a theory like this, but I'm not convinced that it is valid to build the metric from a self-tensor-product like that.  My reasoning is this: take a local orthonormal coordinate frame in which the $A$ 4-vector lies along the $x^0$ direction.  Then $A_1$, $A_2$, and $A_3$ are all zero, so when you take the tensor product, the resulting metric will only have one non-zero component, ie. it will be degenerate.  Maybe if it were only one point that would be fine, but as I see it this will happen at every point in the spacetime.
Am I missing something?  In general the authors seem to be well versed in tensor calculus and so forth, and it is certainly an alluring hypothesis.
 A: OP is correct, this object, $A_μA_ν$, could not be the pseudo-Riemmanian  metric of the usual formulation of electromagnetism in special or general relativity. However, I would not be so quick to completely dismiss the theory without studying it in more details.
First, within the more established way of geometrization of electomagnetism, Kaluza–Klein theory, the object $A_μA_ν$ does crop up as a part of the metric. Specifically: the 5-dimensional metric when restricted to 4-dimensions is not the inverse of “inverse 5-dimensional metric” restricted in turn to 4-dimensions, and the difference is proportional to $A_μA_ν$.
Second, geometrization of Newtonian gravitation, Newton–Cartan theory uses rank 1 temporal metric, $t_{μν}$ to measure time/temporal lengths. This object could be thought of as a tensor product of covector with itself $t_{μν}=τ_μ τ_ν$. The second metric of the Newton–Cartan theory is rank-3 spatial metric to measure distances/spatial lengths.
So, without reading the paper I would not be surprised that one can indeed use $A_μA_ν$ as some sort of a metric in some geometric formulation of electromagnetism.

I am really keen on the idea of geometrizing all of physics using the spacetime manifold …

As an advise, I would recommend to familiarize yourself with more traditional routes of geometrizing physics (such as mentioned above Kaluza–Klein and Newton–Cartan theories) before studying more radical/lesser known ideas.
A: No, here's why. The outer product of a vector with itself is a projection operator with a scaling of the vector's Euclidian length squared, and it projects down to the space parallel to the vector. It is not, therefore, invertible. The metric tensor must be invertible, so a single outer product of a vector with itself cannot be the metric tensor.
You could produce a metric tensor by summing the outer product of 4 linearly independent 4-vector fields with themselves, but that is basically the vielbein formalism.
