I am trying to find out the Eisenstein integers that have norm values of 3,5,7.... I want to see if there is some pattern.
For the norm value of 3, I was able to find six Eisenstein integers. For the norm value of 7, I was able to find twelve Eisenstein integers.
I am unable to find even a single Eisenstein integer with a norm value of 5. Is this correct? Do Eisenstein integers with a norm value of 5 not exist or am I missing something? Does it have to do anything with the fact that 5 is a prime Eisenstein integer?
You are not missing anything. There are no Eisenstein integers
$(a+b\omega$ with $a,b\in\mathbb Z$ and $\omega$ a primitive cube root of $1)$ with a norm of $5.$
One way to see this is to note that $N(a+b\omega)=(a+b\omega)(a+b\overline\omega)$ $=a^2-ab+b^2$
$\equiv a^2+2ab+b^2$ $=(a+b)^2\not\equiv2\pmod3,$ as shown here.
Another way to see this, as suggested in comments by Anne Bauval, is to note that
$N(a+b\omega)=a^2-ab+b^2=\left(a-\frac12b\right)^2+\frac34b^2=5\iff(2a-b)^2+3b^2=20,$
which has no solutions in integers, because $20, 20-3\cdot1^2,$ and $20-3\cdot2^2$ are not squares
and $20-3\cdot b^2<0$ for integers $b>2$.
Finally, if we had $N(a+b\omega)=(a+b\omega)(a+b\overline\omega)=5$,
then $5$ would not be a prime Eisenstein integer.