Let $S_n$ be a random walk process defined by $$S_n=X_1+\dots+X_n$$ with $X_i \sim N(\mu,\sigma^2)$ and $X_i$ are i.i.d.
I'm trying to prove that the quantity $(S_n-n\mu)^2-n\sigma^2$ is a martingale, that is:
$$\Bbb E \bigl[( S_{n+1}-(n+1)\mu)^2-(n+1)\sigma^2 \mid \mathcal{F_{t}} \bigr]=(S_{n}-n\mu)^2-n\sigma^2$$ where $\mathcal{F_{t}}$ is the sigma-algebra generated by $(X_1,\dots,X_n)$. My strategy is based on the extension of another related proof. Showing that $S_n-n\mu$ is a martingale.
We can write $S_{n+1}$ as: $$S_{n+1} = S_n+X_{n+1}$$ and taking the expectation of both sides and conditioning on $S_n$, we have \begin{align*} \Bbb E \bigl[(S_{n+1}\mid S_n\bigr] &=\Bbb E \bigl[S_n+X_{n+1}\mid S_n \bigr]\\ &=S_n+\Bbb E \bigl[X_{n+1}\mid S_n \bigr]\\ &=S_n+\Bbb E \bigl[X_{n+1}\bigr]\\ &=S_n+\mu, \end{align*} where the third step follows because $X_{n+1}$ is independent of $S_n$. Subtracting $(n+1)\mu$ from both sides of the equation: \begin{align*} \Bbb E \bigl[(S_{n+1}-(n+1)\mu \mid S_n\bigr] &=S_n+\mu(1-(n+1))\\ &=S_n-n\mu\\ \end{align*} we have the claim.
Going back to the case of interest, we have that: $$S^2_{n+1} = S^2_n+X^2_{n+1}+2 S_n X_{n+1}$$ Taking the expectation of both sides and conditioning on $S^2_n$, we have: \begin{align*} \Bbb E \bigl[S^2_{n+1}\mid \mathcal{F_{t}} \bigr] &=\Bbb E \bigl[S^{2}_{n}+X^2_{n+1}+2 S_n X_{n+1}\mid \mathcal{F_{t}}\bigr]\\ &=S^2_{n}+\mathbb{E}\big[X^2_{n+1}\mid \mathcal{F_{t}}\big]+2 S_n\mathbb{E}\big[X_{n+1}\mid \mathcal{F_{t}}\big]\\ &=S^2_{n}+\mathbb{E}\big[X^2_{n+1}\big]+2 S_n\mathbb{E}\big[X_{n+1}\big]\\ &=S^2_{n}+\sigma^2+2 S_n \mu\\ \end{align*}
At this point, I'm stuck because I don't know which quantity I should subtract from both sides to complete the proof. For sure, I should subtract $(n+1)\sigma^2$, \begin{align} \mathbb{E}\bigl[S^2_{n+1}-(n+1)\sigma^2\mid S^{2}_{n}\bigr] &=S^{2}_{n}+\sigma^2(1-(n+1))+2 S_n \mu\\ &=S^{2}_{n}-n\sigma^2+2 S_n \mu\\ \end{align} but I'm still far from closing the proof, because I need to subtract other elements that I'm not able to identify.
Any help or hint would be appreciated.
Let me first remark that the notation $\mathcal{F}_t$ doesn't make sense at all. Instead of $\mathcal{F}_t$ you should use $\mathcal{F}_n$, i.e.
$$\mathcal{F}_n = \sigma(X_1,\ldots,X_n)$$
since the right side depends on $n$ and not on $t$.
Moreover, your proof of $S_n-n \cdot \mu$ being a martingale is not correct; you have to replace the conditional expectiation on $S_n$ by the conditional expectation on $\mathcal{F}_n$. (Note that $\sigma(S_n)$ does not define a filtration; $S_n -n \cdot \mu$ is a martingale with respect to $\mathcal{F}_n$.)
Solution 1 Let $(M_n,\mathcal{F}_n)_{n \in \mathbb{N}}$ a martingale such that $M_n \in L^2$. Then we have
$$\begin{align*} \mathbb{E}(M_{n+1}^2 \mid \mathcal{F}_n) &= M_n^2 + \mathbb{E}(M_{n+1}^2-M_n^2 \mid \mathcal{F}_n) \\ &= M_n^2 + \mathbb{E}((M_{n+1}-M_n)^2 \mid \mathcal{F}_n) \\ &= M_n^2 + (A_{n+1}-A_n) \tag{1} \end{align*}$$
where $$A_n := \sum_{k=1}^n \mathbb{E}((M_n-M_{n-1})^2 \mid \mathcal{F}_n)$$ This shows that $M_n^2 - A_n$ is a martingale. $A$ is called compensator of $M$.
In order to apply this general result, we set
$$M_n = S_n - n \cdot \mu$$
and obtain
$$\begin{align*} A_n &= \sum_{k=1}^n \mathbb{E}((S_{n}-(n) \cdot \mu - S_{n-1}+(n-1) \cdot \mu)^2 \mid \mathcal{F}_n) \\ &= \sum_{k=1}^n \mathbb{E}((X_{n}-\mu)^2 \mid \mathcal{F}_n) \\ &= \sum_{k=1}^n \underbrace{\mathbb{E}((X_{n}-\mu)^2)}_{\sigma^2} = n \cdot \sigma^2\end{align*}$$
Consequently, we see that $M_n^2-A_n = (S_n-n \cdot \mu)^2-n \cdot \sigma^2$ is a martingale.
Solution 2 If you prefer straight-forward calculations:
$$\begin{align*} &\quad \mathbb{E}((S_{n+1}-(n+1) \cdot \mu)^2-(n+1) \cdot \sigma^2 \mid \mathcal{F}_n) \\ &= \mathbb{E}\bigg[ ((S_n-n \cdot \mu)+(X_{n+1}-\mu))^2 \mid \mathcal{F}_n \bigg] - (n+1) \cdot \sigma^2 \\ &= (S_n-n \cdot \mu)^2 + 2 (S_n-n \cdot \mu) \cdot \underbrace{\mathbb{E}(X_{n+1}-\mu \mid \mathcal{F}_n)}_{\mathbb{E}(X_{n+1}-\mu)=0} + \underbrace{\mathbb{E}((X_{n+1}-\mu)^2 \mid \mathcal{F}_n)}_{\mathbb{E}((X_{n+1}-\mu)^2) = \sigma^2} - (n+1) \cdot \sigma^2 \\ &= (S_n-n \cdot \mu)^2- n \cdot \sigma^2\end{align*}$$
using that $(S_n-n \cdot \mu)$ is $\mathcal{F}_n$-measurable and $X_{n+1}$ is independent of $\mathcal{F}_n$ with mean $\mu$ and variance $\sigma^2$.