I have $E[x(t)^2]\leq A\operatorname{exp}(Bt)+C/Bt$ it's clear that for a finite time less than $T$, x(t)^2 is a "local martingale" because $\lim E[x(t)^2]<\infty$. But one can see that if $t$ tends to plus infinity then the limit of $E[x^2]$ is also infinity, so this is not a martingale, and we can not apply the convergence theorem.
*Can one tell me if my analysis is right, or if there is any way to extend the local martingale to a martingale to be able to apply the martingale convergence theorem.
*Is there any relation between $x(t)^2$ and $x(t)$ in sens of martingales.
I begin by answering your last question.
Yes, if $X $ is continuous square integrable martingale then $(X_t^2 -\langle X \rangle_t)$ is a continuous martingale.
You are write about not be abble to conclude that $(X^2_t)_{t\geq 0}$ is a martingale since you can not apply the convergence theorem to conclude that $X^2_ t \in L^1(\Omega, \mathcal F_t, \mathbb P) , \ \forall t \geq 0$ (in a filtred probability space $(\Omega, \mathcal F, \mathcal F_t, \mathbb P)$.
However the fact that the $RHS$ of you estimation goes to infinity when $t$ goes to zero or infinity doesn't mean $X^2$ do so.