Proving the relation $\det(I + xy^T ) = 1 + x^Ty$

Hint: Decomposing $$ \begin{pmatrix} 1 & -y^T\\ x & I \end{pmatrix} $$ as lower $\cdot$ upper and upper $\cdot$ lower gives $$ \begin{pmatrix} 1 & 0\\ x & I\end{pmatrix} \cdot \begin{pmatrix} 1 & -y^T\\ 0 & I + xy^T \end{pmatrix} = \begin{pmatrix} 1 + x^Ty & -y^T\\ 0 & I\end{pmatrix} \cdot \begin{pmatrix} 1 & 0\\ x & I \end{pmatrix}. $$

You can also apply the property

$$\det\begin{pmatrix} A & B\\ C & D \end{pmatrix}=\det(A) \det(D-C A^{-1}B)=\det(D) \det(A-B D^{-1}C)$$

(valid when the inverses exist) to the matrix $$\begin{pmatrix} I & -y\\ x^T & 1 \end{pmatrix}$$

(BTW: here I'm assuming column vectors; the question speaks of "column vectors", but its notation would correspond to row vectors)

Here is an approach by upper triangularization for the sake of variety.

Note that $xy^T\in M_n(K)$ has rank $\leq 1$ and $\mbox{tr } xy^T=\sum_{j=1}^nx_jy_j=x^Ty$.

So $0$ is an eigenvalue of multiplicity $n-1$ if $xy^T$ has rank $1$, or $n$ if $xy^T=0$. In the latter case $x=0$ or $y=0$ whence $\det(I+xy^T)=\det I=1=1+x^Ty$. Now if $xy^T$ has rank $1$, take any vector not in $\ker xy^T$ and add it to a basis of $\ker xy^T$ to get a basis of $K^n$. Then, taking the trace to determine the lower-right coefficient, we see that $xy^T$ is similar to $$ S(xy^T)S^{-1}=\pmatrix{0&*\\0&x^Ty}\quad \Rightarrow\quad S(I+xy^T)S^{-1}=\pmatrix{I&*\\0&1+x^Ty} $$ The result follows immediately.

Note: every matrix of rank $1$ is of the form $xy^T$ with $x\neq 0$ and $y\neq 0$. With the approach above, we see that a square rank $1$ matrix is diagonalizable if and only if its trace is nonzero.