A quadratic form is a homogeneous polynomial of degree two, with coefficients in some field. I will write the polynomial as $Q(x)$, where $x$ is vector of length $d$. By some version of Taylor's theorem $$ Q(x+y) = Q(x) +Q(y) + x^T By $$ where $B$ is a $d\times d$ matrix and $x^T By$ is a bilinear form. Since $x+y=y+x$, the form is symmetric, so $x^TBy=y^TBx$ and therefore $B$ is symmetric. The set of points $x$ such that $Q(x)=0$ is a projective quadric, and when $d=3$ it is a projective conic. Over the reals and complexes, $B$ determines $Q$. (And $B$ is what you call the underlying matrix.)
Any proof you have for the real case that a conic is irreducible if $\det(B)\ne0$ will work in the complex case provided it does not use eigenvectors.
Now suppose our field is $\mathbb{C}$. Then $B$ is a complex symmetric matrix, but we can say very little about how to diagonalize a symmetric matrix over $\mathbb{C}$, we can only unitarily diagonalize a matrix if it is normal ($BB^*=B^*B$) and a complex symmetric matrix need not be normal.
We do not call the matrix $B$ a quadratic form, because it is a matrix and not a polynomial.
If people talk about a conic over $\mathbb{C}$, then we assume that the coefficients of the polynomial and the entries of $B$ may be complex. This is standard.
Using eigenvalues and eigenvectors makes it easier to work with quadratic forms over the reals. However this approach does not work over $\mathbb{C}$ or over finite fields.
(Edit. In response to the question below.) Since $Q(x)$ is quadratic, if it factors then it must be a product of two linear factors, so we can then write it as $(c^Tx)(d^Tx)$ (where $c$ and $d$ could be parallel). Now we can calculate $$ Q(x+y) -Q(x) -Q(y) = (c^Tx)(d^Ty)+(d^Tx)(c^Ty). $$ If we set $B = cd^T+dc^T$ (which is the sum of two $d\times d$ matrices, then $$ x^TBy = (x^Tc)(d^Ty) +(x^Td)(c^Ty) = (c^Tx)(d^Ty)+(d^Tx)(c^Ty). $$ So the matrix of the quadratic form is $cd^T+dc^T$. Since we are in three dimensions (over $\mathbb{C}$ or $\mathbb{R}$ it does not matter), there is a non-zero vector $h$ say such that $c^Th=d^Th=0$. Therefore $Bh=0$ and $B$ is not invertible.
For the other direction, if $\det(B)=0$ then $B$ has rank at most two. Now if $B$ has rank two and is symmetric we can write it as $B=cd^T+dc^T$ for two vectors $c$ and $d$. (This is proved in my book with Royle, it is reasonably standard linear algebra. There might be a way to simplify this but my dinner's calling...) Now since $$ 4Q(x) = Q(2x) =Q(x+x) = Q(x)+Q(x) +x^TBx $$ we see that $x^TBx=2Q(x)$. So $2Q(x)$ factors, because $x^TBx=2(c^Tx)(d^Tx)$.
This is probably a bit dense, but you can see that we are not using properties of the field, other than it does not have even characteristic.