Do polynomials in two variables always factor in linear terms?

My candidate to fail is $xy+1$. If this is the product of linear terms, we can have at most two of degree one. So $$ xy+1=(ax+by+c)(dx+ey+f). $$ We need $ad=be=0$, to avoid the squares. If $a=0$, then $b\ne0$ and so $e=0$. We have $$ xy+1=bdxy+bfy+cdx+cf. $$ Then $ cf=1$, $bf=cd=0$. So $b=d=0$, making the equality impossible. The other case is similar.

Let me add to Martin's perfect answer that a homogeneous polynomial $f(x,y)$ (=sum of monomials of the same degree) in two variables does factor into linear homogeneous factors in an essentially unique way, that is up to permutations of the factors and multiplication of the factors by constants. More explicitly: $$f(x,y) =\sum_{i+j=d} a_{ij}x^iy^j=\prod _{k=1}^d(u_ix+v_iy)$$ However this is no longer true if the number of variables is $\geq 3$.
For example the polynomial $x^2+y^2+z^2$ is irreducible i.e. has no non-trivial factorization.

Finally note that if a homogeneous polynomial has a factorization, then the factors must be homogeneous too. For example : $$x^3+y^3+z^3-3xyz=(x+y+z)(x+wy+w^2z)(x+w^2y+wz) $$ (where $w=e^{2i\pi/3} $)

Edit
By request of a great contributor to this site, here is an explanation of why a homogeneous polynomial $f(x,y)$ in two variables over an algebraically closed field can be factored into linear polynomials:
Just write $$f(x,y)=\sum _{i=0}^da_{i}x^iy^{d-i}=y^d\sum _{i=0}^da_i(\frac xy)^i=y^d\prod_{i=1}^d (u_i(\frac xy)+v_i)=\prod_{i=1}^d (u_ix+v_iy)$$ The penultimate equality results from the factorization of a univariate polynomial into degree one factors.
[The calculation above is true only if $a_d\neq0$, i.e. if $f$ is not divisible by $y$.
If $f$ is divisible by $y$, one must very slightly modify the above but $f$ is still a product of linear factors, some of them now being equal to $y$. For example $x^2y^2+y^4=y^2(x+iy)(x-iy)$ ]

To show that this is not the case, look at $x^2 + y^2 - 1 = 0$, which is true all along the unit circle, if your idea was true that we could factor down to "individual zero points", we would need to have infinite number of factors since there are infinitude of points on the unit circle. Also linear terms would not be possible, since the unit circle is curved and not a set of lines.