The concept of norm is extremely important, including the fact that it is multiplicative (meaning that $N(m) N(n) = N(mn)$). Learn it, love it, live it.
What really makes things easy in $\mathbb{Z}[\sqrt{-6}]$ (which is the same thing as $\mathbb{Z} + \mathbb{Z}\sqrt{-6}$, right?) is the fact that there are only two units: 1 and $-1$. How do we know those are the only units? A unit has norm 1 or $-1$. The norm of a number of the form $a + b \sqrt{-6}$ is $a^2 + 6b^2$, so the norm can't be negative. The possible norms in $\mathbb{Z}[\sqrt{-6}]$ are 0, 1, 4, 6, 7, 9, 10, etc. (see Sloane's A002481). Both 1 and $-1$ have a norm of 1, no number in this domain can have a norm of $-1$. If $a > 1$ or $b > 1$, then $N(a + b \sqrt{-6}) > 1$.
2 and 3 are potentially prime because no number in this domain has a norm of 2 or 3. But, as you already know from the other question you asked, $2 \times 3 = -(\sqrt{-6})^2 = 6$, which means 2 and 3 are irreducible but not prime in this domain.
But what about $\sqrt{-6}$? Its norm is 6. If it's reducible, we can find two numbers, neither of them 1, such that $N(\alpha) N(\beta) = 6$. So we're looking for a number with a norm of 2 and another with a norm of 3. But we just saw that there are no numbers in that domain with that norm. Therefore, $\sqrt{-6}$ is irreducible.
If you really want to be sure, you can try each product $(a - b \sqrt{-6})(a + b \sqrt{-6})$ for $0 \leq a \leq 3$ and $b$ likewise.
Yes, you are on the right track. All your reasoning makes sense to me.
On your question about the associates
By the properties of the norm, associates have the same norm. So the only possible associates in your list are $1 + \sqrt{-5}$ and $1 - \sqrt{-5}$.
Now determine all units of $\mathbb{Z}[\sqrt{-5}]$ by finding all integer solutions of $N(a + b\sqrt{-5}) = 1$. (The list will be quite short.)
Then check every unit $u$ for $u (1 + \sqrt{-5}) = (1 - \sqrt{-5})$.
You will see that this never happens and thus, the two elements are not associate.
Best Answer
If $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{D}]$, is then $|\mathsf{N}(\alpha)| = p$ certain to be a prime number in $\mathbb{Z}$? (I'm stating the converse to make sure we're on the same page; if I've misunderstood what the converse is here then we'll never hear the end of it).
No. Consider $D = -10$. It turns out that $\sqrt{-10}$ is irreducible, yet it has a norm of 10. The formula for norm in this domain works out to be $$\mathsf{N}(a + b\sqrt{-10}) = a^2 - (-10)b^2 = a^2 + 10b^2.$$ That's all a norm is: a function that enables you to compare numbers from other domains within the familiar framework of $\mathbb{Z}^+ \bigcup \{0\}$. The Euclidean algorithm is another familiar thing from $\mathbb{Z}$, but, unlike the norm, it can't always be carried over.
The norm can never be a negative number here, so there's no need to specify absolute value (quite a different story if $D$ is positive). If $\sqrt{-10}$ is reducible, then we can solve $\mathsf{N}(\beta) = 2$ and $\mathsf{N}(\gamma) = 5$. Except we can't. The possible norms in this domain are 0, 1, 4, 9, 10, 11, 14, 16, 19, 25, ... (see Sloane's http://oeis.org/A020673). Either $\beta$ or $\gamma$ is a unit.
Two prior answerers already mentioned the norms of rational primes. I mention it again for the sake of completeness.