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.