I thank Snoop for a comment and John Dawkins for his other answer which led me to elaborate a bit to show what can go wrong (and in fact was going wrong in my earlier versions):
Since the $X_k$ are i.i.d.
\begin{align}\tag{1}
\mathbb E[X_k^2]&={\rm Var}[X_1]+1=:\sigma^2+1\,.
\end{align}
The quadratic variation is the unique increasing predictable
process $\langle M\rangle_n$
such that $M_n^2-\langle M\rangle_n$ is a martingale. For a discrete process predictable means that at $n$ it has to be ${\cal A}_{n-1}$-measurable.
Notation:
\begin{align}
A_n&=\sum_{k=1}^{n-1}\sigma^2M_k^2\,,\tag{2}\\
B_n&:=\sum_{k=1}^n(M_k-M_{k-1})^2\,,\tag{3}\\
C_n&:=\mathbb E\Big[B_n\Big|{\cal A}_{n-1}\Big]\tag{4}
\end{align}
When $\sigma>0$ only the two processes $M_n^2-A_n\,,M_n^2-B_n$ are martingales, and only $A_n$ is the predictable quadratic variation $\langle M\rangle_n\,$.
$B_n$ is not predictable.
$C_n$ is predictable but not necessarily increasing and does not make $M_n^2-C_n$ a martingale.
Proof.
Using the i.i.d properties of the $X_i\,,$
\begin{align}
\mathbb E[M_{n+1}^2|{\cal A}_n]&=\mathbb E\Big[\prod_{k=1}^{n+1}X_k^2\,\Big|{\cal A}_n\Big]=
\mathbb E[X_{n+1}^2]\prod_{k=1}^nX_k^2=\sigma^2M^2_n+M^2_n\,.\tag{5}
\end{align}
Since taking the conditional expectation twice does not change the result,
\begin{align}
\mathbb E\Big[C_{n+1}\,\Big|{\cal A}_n\Big]&=\mathbb E\Big[B_{n+1}\,\Big|{\cal A}_n\Big]=\underbrace{\mathbb E\Big[(M_{n+1}-M_n)^2\,\Big|{\cal A}_n\Big]}_{(*)}+B_n\,.\tag{6}
\end{align}
Because $M_n$ is a martingale we can further develop the (*) term in (6):
\begin{align}
(*)&=\mathbb E\Big[M_{n+1}^2 \,\Big|{\cal A}_n\Big]-2M_n^2+M_n^2
\stackrel{(5)}{=}\sigma^2M^2_n\,.\tag{7}
\end{align}
Therefore, by (5) and (7),
$$
\mathbb E\Big[M_{n+1}^2-C_{n+1}\,\Big|{\cal A}_n\Big]=M^2_n-\color{red}{B_n}\,,\tag{8}
$$
that is, $M_n^2-C_n$ is $\color{red}{\rm not}$ a martingale.
The proof that $M_n^2-B_n$ is a martingale is the same. Use (5)-(7).
To see that $M_n^2-A_n$ is a martingale we only have to observe that, since $A_{n+1}$ is ${\cal A}_n$-measurable by definition,
\begin{align}
\mathbb E\Big[A_{n+1}\,\Big|{\cal A}_n\Big]&=\sigma ^2M_n^2+ A_n\tag{9}
\end{align}
trivially holds. Combining (5) with (9) shows that $M_n^2-A_n$ is a martingale.
$$\tag*{$\Box$}
\quad
$$
Remark.
The situation in the discrete case is a bit analogous to discontinuous processes. We know that
for the Poisson process $N$ the predictable quadratic variation is $\langle N\rangle_t=\lambda t\,.$ But there is also the quadratic variation
$$
[N]_t=\sum_{s\le t}(N_s-N_{s-})^2=\sum_{s\le t}(\Delta N_s)^2=\sum_{s\le t}\Delta N_s=N_t\tag{10}
$$
which makes $N_t-[N]_t$ a martingale. But $[N]_t$ is not predictable.
