Trivial norm is equivalent only to itself (on a field)
The norm here fulfills $3$ requirements
$(1) |x|\ge0$ and $ |x|=0\iff x=0$
$(2)|x\cdot y|=|x|\cdot|y|$
$(3)|x+y|\le|x|+|y|$
$2$ norms are said to be equivalent iff they induce equivalent metrics, and $2$ metrics $d_1,d_2$ are said to be equivalent if a sequence is Cauchy w.r. to $d_1$ iff it is Cauchy w.r. to $d_2$
So when I pick the trivial norm, i.e. $ |x| = \left\{ \begin{array}{ll} 1 & \mbox{if $x \neq 0$};\\ 0 & \mbox{if $x = 0$}.\end{array} \right. $
and any nontrivial norm $|.|'$ then choose any nonzero element $x$ with $|x|'\neq1$ such an element must exist so if $|x|'>1$ then take $x^{-1}=:y$ so $|y|'<1$ and the sequence $\{y^n\}_n$ is Cauchy w.r. to $|.|'$ because for $n\ge m\ge N:|y^n-y^m|\le|y^n|+|y^m|\le2|y^m|\le2|y^N|$ and not Cauchy in $|.|$ because $\nexists N;n,m\ge N\implies|y^n-y^m|<1$
Is this enough ? I cannot mention any topology or balls, I have to stick to the definition
I think your argument is essentially correct. You can add to clarify that in the trivial norm, $\lvert z \rvert < 1 \Leftrightarrow z = 0$ and a sequence is Cauchy if and only if it is eventually constant. Also, the last line should read $\not \exists N: \color{red}{\forall} m,n \ge N: \lvert y^m-y^n\rvert < 1$ which you still have to show -- I would write $y^m - y^n = y^m (y^{m-n}-1)$, the first factor cannot be $0$ because $y \neq 0$ and we're in a field, the second factor cannot be zero for $m \neq n$ because $\lvert y^{m-n} \rvert = \lvert y \rvert^{m-n} \neq 1 = \lvert 1 \rvert$ by choice of $y$.
Besides those (maybe pedantic) mathematical points, I think your argument would also be improved by correct sentences, grammar, and punctuation. That is no small issue. As it stands, your decisive paragraph is a run-on sentence going "So when ... then ... and ... because ... and not ... because ... ." Please break that up into smaller bits.