Irreducibility of polynomials in two variables
A trick which works surprisingly often in my experience: If the Newton polytope of $f$ can not be written as a Minkowski sum of two smaller polytopes, then $f$ is irreducible. I think of this as a generalization of Eisenstein's criterion.
It is surprisingly easy to test whether a lattice polygon in $\mathbb{R}^2$ can be written as a Minkowski sum of smaller lattice polygon. Let $P$ be a lattice polytope. Travel around $\partial P$ and write down the vectors pointing from each lattice point to the next lattice point; call this sequence of vectors $v(P)$. For example, if our polynomial is $a y^2 + b y + c xy + d + e x + f x^2 + g x^3$, with $adg \neq 0$, then the lattice points on the boundary are $(0,2)$, $(0,1)$, $(0,0)$, $(1,0)$, $(2,0)$, $(3,0)$ so $v(P) =(\ (0,-1),\ (0,-1),\ (1,0),\ (1,0),\ (1,0),\ (-3,2)\ )$. (Note that $(1,1)$ is not on the boundary of the triangle.)
It turns out that $v(A + B)$ is simply the sequences $v(A)$ and $v(B)$, interleaved by sorting their slopes . So, if $P$ can be written as the Minkowski sum $A+B$, we must be able to partition $v(P)$ into two disjoint sub-sequences, each of which sums to zero. In the above example, this can't be done, so any polynomial of the form $a y^2 + b y + c xy + d + e x + f x^2 + g x^3$, with $adg \neq 0$ is irreducible.
As an example of a polynomial which could factor, look at $a y^2 + by + c xy + dx + e x^2$. So the boundary is $(0,2)$, $(0,1)$, $(1,0)$, $(2,0)$, $(1,1)$ with $v(P) = (\ (0,-1),\ (1,-1),\ (1,0),\ (-1,1),\ (-1, 1)\ )$. This is the interleaving of $(\ (0,-1),\ (1,0),\ (-1, 1)\ )$ and $(\ (1,-1),\ (-1,1)\ )$, so a polynomial with this Newton polytope could factor.
One of my all-time leading candidates for Most Preposterous Theorem Ever:
Definition: A polynomial $f(x) \in \mathbb{C}[x]$ is indecomposable if whenever $f(x) = g(h(x))$ for polynomials $g$, $h$, one of $g$ or $h$ is linear.
Theorem. Let $f, g$, be nonconstant indecomposable polynomials over $\mathbb C$. Suppose that $f(x)-g(y)$ factors in $\mathbb{C}[x,y]$. Then either $g(x) = f(ax+b)$ for some $a,b \in \mathbb{C}$, or $$\operatorname{deg} f = \operatorname{deg} g = 7, 11, 13, 15, 21, \text{ or } 31,$$ and each of these possibilities does occur.
The proof uses the classification of the finite simple groups [!!!] and is due to Fried [1980, in the proceedings of the 1979 Santa Cruz conference on finite groups], following a the reduction of the problem to a group/Galois-theoretic statement by Cassels [1970]. [W. Feit, "Some consequences of the classification of finite simple groups," 1980.]
If $k$ is algebraically closed, then any two components of the projective closure of $\text{Spec } k[x, y]/(f(x, y))$ intersect by Bezout's theorem, and one can check for the existence of such points by looking at where the partial derivatives simultaneously vanish (the singular points). For example, $f(x, y) = x^2 + 2xy + y^2 - 1$ has projective closure defined by $F(X, Y, Z) = X^2 + 2XY + Y^2 - Z^2$ and the partial derivatives vanish at $(1 : -1 : 0)$, the intersection of the components $X + Y - Z = 0$ and $X + Y + Z = 0$.
Generically, $k[x]$ is a UFD, so Eisenstein's criterion applies, although I am not sure how practical this is.