Let $\mathbb R^m\ni \mathbf x \ne \mathbf 0 \ne \mathbf y\in\mathbb R^n$. Let $A = \mathbf x\mathbf y^T$ and find the single non-zero eigenvalues of $A$. Note the compact SVD of this matrix $A$ is
$$\left(\frac x {\|x\|}\right)\cdot (\|x\|\|y\|)\cdot \left(\frac {y^T} {\|y\|}\right) = (U_1)\cdot (S)\cdot(V_1).$$
Note the liberal use of parenthesis to avoid any ambiguity of what $U_1,S,V_1$ are.
Can I proceed similarly to this post to try and find the single $\lambda \ne 0$? See that $Au = xy^Tu = \lambda u$. Then we have $xy^Tu = \lambda u \implies y^Tx(y^Tu) = \lambda y^T u \implies \lambda = y^Tx \equiv (\|y\|\|x\|)^2 \implies \sigma = \sqrt{\lambda} = \|y\|\|x\|$?
It's simply $A x = x y^T x = (y^T x) x$ so $x$ is an eigenvector for eigenvalue $\lambda$.