Using the necessary condition for nilpotent elements - they cannot be invertible, we fix the element of the ring ax+b and try to solve this equation:
"(ax+b)(cx+d)=1\\\\\nacx^2+(ad+bc)x+bd=1\\\\\n(ad+bc)x+(bd-3ac)=1\\\\\n\\begin{cases}\n ad+bc=0\\\\\n bd-3ac=1\n \\end{cases}\\\\\nBecause\\ \\mathbb{Z}_7 \\ is \\ a \\ field\\ from \\ first\\ equation:\\\\\nc=-\\frac{ad}{b}\\\\\nPut\\ in\\ second:\\\\\nbd+\\frac{3a^2d}{b}=1"
It is easy to see that this equation is solvable for b not equal to 0
If b=0:
"ax(cx+d)=1\\\\\nadx-3ac=1"
And this equation is easily solvable for a not equal to 0, which means that there are no nilpotent elements in the ring
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