Is there a systematic way of finding the factorization over the closure of $\mathbb{Z}_2$ of $p(x) = x^{32} - x$? And similar polynomials of the form $x^{p^n} - x$.  I know that the degrees of the irreducible monic polynomials that factorize $x^{32} - x$ will have degree $d \vert 5 = 1, 5$.  Also, I know that $x$ and $(x-1)$ are factors, so that leaves me with: 
$x^5, x^5+x^4, x^5+x^4+x^3, x^5+x^4+x^3+x^2, ..., x^4, x^4+x^3, x^4+x^3+x^2,...$
It seems like I can spend an entire month working by trial and error.  Is there some better way?

It sounds like you know that $x^{32} - x$ is $x(x - 1)$ times all irreducible polynomials of degree $5$. There are $6$ of these out of the $32$ (monic) polynomials of degree $5$. You can rule out most of these by checking for linear factors: first, everything divisible by $x$ is reducible, which leaves $16$ possibilities (constant term must be $1$). Incidentally, this rules out every example you wrote down, some of which aren't even of degree $5$.
Second, everything divisible by $x + 1$ is reducible, which leaves $8$ possibilities (sum of terms must be odd), namely
$$x^5 + x + 1$$
$$x^5 + x^2 + 1$$
$$x^5 + x^3 + 1$$
$$x^5 + x^4 + 1$$
$$x^5 + x^3 + x^2 + x + 1$$
$$x^5 + x^4 + x^2 + x + 1$$
$$x^5 + x^4 + x^3 + x + 1$$
$$x^5 + x^4 + x^3 + x^2 + 1$$
$6$ of these are irreducible, so you want to rule out $2$ more. These $2$ are the ones of the form (irreducible quadratic)(irreducible cubic). There's exactly one irreducible quadratic, namely $x^2 + x + 1$, so the only thing you need to do is check what's divisible by this. $x^2 + x + 1$ divides $x^3 - 1$, so you can reduce the exponents $\bmod 3$ to do this. You get that $x^5 + x + 1$ and $x^5 + x^4 + 1$ are divisible by $x^2 + x + 1$ and that's it; the other $6$ are the irreducibles of degree $5$. 
In total this took less than 5 minutes, I think. 
