I realize this question was asked before, but I did not find the answers satisfying. Here is my attempt:
Since G is cyclic, any element can be written as $g^m$ for $ 0 \leq m \leq 11 $, so the equation reads $ (g^k)^2 = g^j $. These two elements are equal if $ 2k \equiv j \mod 12 $ which implies $j = 2(m - 6 z)$, so if we want to find a $g$, which does not then it has to be an element with $j$ odd.
However, the solution says that $ g^6 = (g^2)^6 = e $, which is a contradiction.