There's a relatively low-level reason, partly mentioned by Hagen von Eitzen, that is $2$ is the smallest prime of them all, and hence appears in more places than others, even in the most abstract theorems. This kind of small special cases turns up everywhere because constants are generally small.
For example, consider the quadratic equation $ax^2 + bx + c = 0$ for $a,b,c \in F$ for some field $F$. One special case is $a = 0$, which we can easily deal with. Then we proceed by the usual complete-the-square technique to get $(2ax+b)^2 = b^2-4ac$. Suppose we let $\sqrt{d}$ denote some arbitrary square root of $d$ in $F$, we then get $2ax = -b \pm \sqrt{b^2-4ac}$ as usual. Since $a \ne 0$, we can divide both sides by $a$, but we're stuck with the $2$ unless $char(F) \ne 2$. Now how did the $2$ appear? It was a completely natural outcome of the identity $(s+t)^2 = s^2 + 2st + t^2$ involved here, simply because $1+1 = 2$. It so happens that $2$ is a prime, and so the characteristic that we need to exclude is $2$ itself to ensure that $2$ has a multiplicative inverse.
Similarly, when solving the cubic equation in an arbitrary field, one finds exactly the same issue with $3$. If we did it by Lagrange resolvents after obtaining the depressed cubic, we would have three roots $a,b,c$ such that $0 = a+b+c$, and we would let $x = a + ζ b + ζ^2 c$ and $y = a + ζ^2 b + ζ c$ where $ζ$ is a primitive cube root of unity. It turns out that $x^3,y^3$ are both in $F(\sqrt{D})$ where $D$ is the discriminant of the cubic, and so we can express them in terms of $a,b,c$ alone, provided $2$ has an inverse ($2$ appears in the computation 'by accident'). Then we add all three equations and divide by $3$ to get $a$, which of course needs $3$ to be invertible. So at least for this method we need the field to have characteristic neither $2$ nor $3$.