(*N.B.: I have modified this answer to take into account the comments below about the case of characteristic $2$, when in fact, the discriminant is the square of an irreducible polynomial.*)

The discriminant is defined by the property that, when
$$
p(x) = x^d - s_1 x^{d-1} + \cdots + (-1)^ds_d = (x-t_1)(x-t_2)\cdots(x-t_d)
$$
(i.e., when one substitutes the $i$th elementary symmetric function of the $t_k$ for $s_i$), the discriminant becomes
$$
D(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j)^2.
$$
Suppose there were a factorization $D(s_1,\ldots,s_d) = D_1(s_1,\ldots,s_d)D_1(s_1,\ldots,s_d)$ where each $D_i$ had positive degree.  Then by unique factorization, when one substitutes as above, one must be able to write, for $a = 1,2$,
$$
D_a(s_1,\ldots,s_d) = c_a \prod_{(i,j)\in S_a} (t_i-t_j)
$$
where $c_1$ and $c_2$ are (nonzero) constants and $S_1$ and $S_2$ are disjoint nonempty subsets of the set of pairs $(i,j)$ in $\{1,\ldots,d\}$ whose union is the entire set of distinct pairs in this set.   Thus, for example, $D_1(s_1,\ldots,s_d)$ will vanish when $t_i=t_j$ (for $i\not=j)$ only if $(i,j)$ or $(j,i)$ belongs to $S_1$.  Since one can't detect which of the $t_i$ are equal using only $s_1,\ldots, s_d$, it follows that $S_1$ must contain either $(i,j)$ or $(j,i)$ for each distinct pair.  The same argument applied to $D_2$ shows that $S_2$ also must contain either $(i,j)$ or $(j,i)$ for each distinct pair.  Thus, one must have, for $a = 1,2$,
$$
D_a(s_1,\ldots,s_d) = c'_a \prod_{i<j} (t_i-t_j)
$$
for some constants $c'_1$ and $c'_2$.  

However, *when the characteristic of the field is not $2$*, one cannot have a polynomial $E(s_1,\ldots,s_d)$ such that, after substitution, one obtains
$$
E(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j)
$$
since the left hand side cannot detect permutations in the $t_i$, whereas the right hand side will change sign when one makes an odd permutation of the $t_i$.
Thus, $D$ is irreducible when the characteristic of the field is not $2$.

As Jarek Kuben pointed out in the comments below, when the characteristic of the field is $2$, the expression
$$
F = \prod_{i<j} (t_i-t_j) = \prod_{i<j} (t_i+t_j)
$$
is symmetric in the $t_i$, so $F$ can be written as a polynomial in the $s_i$.  One then has $D = F^2$, so $D$ is a square.  The above argument shows, however, that $F$ must be irreducible, since any factorization $D = D_1D_2$ has to have $D_1 = D_2 = F$.