Here it is what I have progressed:
$Z_{t}\overset{def}{=}log(1+M^{2}_{t})$, Then there follows: $E(Z_t|\mathcal{F}_{s})=E\Big[log(1+M^{2}_{t})|\mathcal{F}_{s}\Big]\leq log\Big[E(1+M^{2}_{t}|\mathcal{F}_{s})\Big]=log\Big[1+E{(M^{2}_{t}|\mathcal{F}_{s}})\Big]$, for $s\leq t$. The inequality "$\leq$" here is a simple result given by Jensen's inequality.
Suppose that $E(M^{2}_{t}|F_{s})=M^{2}_{s}$ could hold, then $E(Z_{t}|F_{s})\leq log(1+M^{2}_{s})=Z_{s}$ would hold, which means that $Z_{t}$ would be a supermartingale. However, a squared martingale is indeed a submartingale, again because the Jensen's inequality holds.
So what I have concluded so far is only that $Z_{t}$ is not a supermartingale.
