I can prove that it is true when $A$ is a field. When $A$ is a field any $\langle X\rangle$ is a maximal ideal of $A[X]$. Any maximal ideal of $A[X]/\langle X^n\rangle$ is of the form $M/\langle X^n\rangle$ for some maximal ideal $M$ of $A[X]$ containing $X^n$. As $M$ is maximal, $M$ is prime and $X^n\in M\implies X\in M\implies\langle X\rangle \subseteq M\implies M=\langle X\rangle$. This proves $\langle X\rangle/\langle X^n\rangle $ is the only maximal ideal of $A[X]/\langle X^n\rangle$.
But I'm unable to conclude anything for any local ring $A$. Can anyone help complete the solution? Thanks for your help in advance.
As you said, a given maximal ideal corresponds to $M/(X^n)$ where where $M\lhd A[X]$ is a maximal ideal. Then as you discovered $(X)\subseteq M$. And since $A[X]/(X)\cong A$, the maximal ideals of the former correspond to those of the latter. If $A$ is local, then there is only one maximal ideal of the latter ring, hence only one maximal ideal of the former ring.