As far as I know there is no way of finding the discriminant without finding the full ring of integers in the process.
Indeed, once you know the ring of integers, finding the discriminant is a trivial piece of linear algebra; and conversely, if you know the discriminant in advance, that makes finding the ring of integers much easier (because once you've found enough integers to generate a subring with the right discriminant, you know you can stop). So determining the discriminant cannot be all that much easier than determining the integers.
As for how to find the integers, it is a very well-studied problem; if you're interested in such things, Henri Cohen's "A course in computational algebraic number theory" is a very good reference.
I will write $t$ instead of $\theta$. So we work in the field $K=\Bbb Q[T]/(T^3+T+1)=\Bbb Q[t]$, where $t$ is the image of $T$ modulo the ideal $(T^3+T+1)$. Consider the basis as a $\Bbb Q$ vector space of $K$ given by the algebraic integers
$$
1,\ t,\ t^2\ .
$$
We fix an embedding of $K$ into $\Bbb C$, let $t_1$ be the image of $t$.
The polynomial $X^3+X+1\in K[X]$ has three roots in $\Bbb C$, one is $t_1$, and let $t_2,t_3$ be the other two. There are thus three embeddings $\sigma_k:K \to \Bbb C$, given by $\sigma_k:t\to t_k$. The discriminant of the basis $1,t,t^2$ is now (in my very ad-hoc computation):
$$
\begin{aligned}
\Delta[1,t,t^2]
&=\det{}^2(\ (\sigma_k(t^j))_{1\le k\le 3; 0 \le j\le 2}\ )\\
&=
\begin{vmatrix}
1&1&1\\
t_1&t_2&t_3\\
t_1^2&t_2^2&t_3^2
\end{vmatrix}^2
\\
&=(\ (t_1-t_2)(t_1-t_3)(t_2-t_3)\ )^2\text{ (Vandermonde)}
\\
&=-(t_1-t_2)(t_1-t_3) \cdot (t_2-t_1)(t_2-t_3)\cdot(t_3-t_1)(t_3-t_2)
\\
&=-\prod(t_1-t_2)(t_1-t_3)=-\prod(t_1^2-(t_2+t_3)t_1+t_2t_3)
\\
&=-\prod(2t_1^2-(t_1+t_2+t_3)t_1+t_2t_3)=-\prod(2t_1^2+t_2t_3)
\\
&=-\prod\frac 1{t_1}(2t_1^3+t_1t_2t_3)
=-\frac 1{t_1t_2t_3}\prod(2t_1^3-1)
\\
&=-\frac 1{-1}\prod(-2t_1-2-1)=-(2t_1+3)(2t_2+3)(2t_3+3)
\\
&=-(27+18\sum t_1+12\sum t_1t_2+8t_1t_2t_3)
\\
&=-(27+0+12-8)=-31\ .
\\[3mm]
&\text{Alternatively one can comuute the resultant}
\\
\Delta[1,t,t^2]
&=\pm\operatorname{Resultant}(\ X^3+X+1\ ,\ (X^3+X+1)'\ )
\\
&=
\pm
\begin{vmatrix}
1 & 0 & 1 & 1 & \\
& 1 & 0 & 1 & 1\\
3 & 0 & 1 & & \\
& 3 & 0 & 1 & \\
& & 3 & 0 & 1
\end{vmatrix}
\end{aligned}
$$
(The sign factor above is $(-1)$ to the power $3(3-1)/2$, so it is $-1$.)
Now we come to your question. We have the discriminant computed for the integral basis $1,t,t^2$, it is $-31$ and it is squarefree. Well, in this case the base is integral. In the hypothetical case of an integral system $(v_1,v_2,v_3)$ with a discriminant $\Delta[v_1,v_2,v_3]\in\Bbb Z$ where there is a prime factor $p$ that appears to the power $\ge $ two in the discriminant, then we may search for an integral linear combination of the shape $\frac 1p(n_1 v_1+n_2v_2+n_3v_3)$, which is guaranteed to exist.
Reference: Computing integral bases, John Paul Cook, see for instance Theorems 3.3. and 3.4.
Best Answer
Let's look at the first part. Let $R$ be the ring of integers of $\mathbb Q[\alpha]$. Let $\mathfrak p$ be a maximal ideal of $R$ containing $2$. Then $\alpha^5=2(1+\alpha)\in \mathfrak p$, hence $\alpha\in \mathfrak p$. Moreover $1+\alpha\notin \mathfrak p$, so $2\in \mathfrak p^5R_\mathfrak p$. This implies that the ramification index at $\mathfrak p$ is at least $5$. But $\mathbb Q[\alpha]$ has degree $5$, so the ramification index is exactly $5$, the residue extension is trivial and there is only one prime above $2$. By Problem 21 just before the one you are considering, $2^4$ divides the discriminant of $R$.
Now the discriminant of $X^5-2X-2$ is $2^4.3.13.67$. So it is equal to the discriminant of $R$ and therefore $\mathbb Z[\alpha]=R$.
The seconde part should be solved similarly.
Edit To see the ramification index at $\mathfrak p$ is at least $5$ without using localization (I don't know how things are organized in Marcus), we can just compare ideal decompositions of $\alpha^5 R$ and $2(1+\alpha)R$, noticing that no prime ideal can contains at the same time $\alpha$ and $1+\alpha$.