Let $(B_{t})_{t\geq 0}$ be a Brownian motion and $(\mathcal{F}_{t})_{t\geq 0}$ its canonical filtration. Find a Stochastic Process $(A_{t})_{t\geq 0}$ such that $(B_{t}^{4}+A_{t})_{t\geq 0}$ is a Martingale.
I don't want to settle for a single $(A_{t})_{t\geq 0}$ process, I would like to be able to find a family of processes.
My attempt: Let $X_{t}:=B_{t}^{4}+A_{t}$ and $t>s$, then we have \begin{align} \mathbb{E}[X_{t}-X_{s}|\mathcal{F}_{s}]&= \mathbb{E}[B_{t}^{4}+A_{t}-B_{s}^{4}-A_{s}|\mathcal{F}_{s}]\\ &=3(t-s)^{2}+6(t-s)B_{s}^{2}+\mathbb{E}[A_{t}-A_{s}|\mathcal{F}_{s}] \end{align} If $(X_{t})_{t\geq 0}$ is a martingale, then we should have $$0=3(t-s)^{2}+6(t-s)B_{s}^{2}+\mathbb{E}[A_{t}-A_{s}|\mathcal{F}_{s}]$$ and and we should also have that $(A_{t})_{t\geq 0}$ is adapted to filtration $(\mathcal{F}_{t})_{t\geq 0}$. Therefore, we have $$A_{s}=3(t-s)^{2}+6(t-s)B_{s}^{2}+\mathbb{E}[A_{t}|\mathcal{F}_{s}] \tag{$\bigstar$}$$
Therefore, it all comes down to solving equation ($\bigstar$).
(I take it that you are looking for a non-trivial solution - otherwise you could simply take $A_t := -B_t^4$, as was pointed out in a comment.)
Using that $(B_t)_{t \geq 0}$ has independent and stationary increments and the fact that $B_t \sim N(0,t)$, it follows that
$$M_t^{(\lambda)} := \exp \left(\lambda B_t - \frac{1}{2} \lambda^2 t \right)$$
is a martingale for any $\lambda \in \mathbb{R}$. Equivalently,
$$\int_F M_s^{(\lambda)} \, d\mathbb{P} = \int_F M_t^{(\lambda)} \, d\mathbb{P} \tag{1}$$
holds for all $s \leq t$ and $F \in \mathcal{F}_s$. Now the idea is to differentiate $(1)$ with respect to $\lambda$. For the first derivative, we get
$$\int_F (B_s-\lambda s) M_s^{(\lambda)} \, d\mathbb{P} = \int_F (B_t-\lambda t) M_t^{(\lambda)} \, d\mathbb{P}. \tag{2}$$
Since this holds in particular for $\lambda=0$, we see that $\int_F B_t \, d\mathbb{P} = \int_F B_s \, d\mathbb{P}$ for all $s \leq t$ and $F \in \mathcal{F}_s$, which means that $(B_t)_{t \geq 0}$ is a martingale (no surprise here; we know this already). By calculating higher derivatives, we can increase the "power" of the martingale. E.g. by differentiating $(2)$ once more again with respect to $\lambda$, we find that
$$\int_F ((B_s-\lambda s)^2-s) M_s^{(\lambda)} \, d\mathbb{P} = \int_F ((B_t-\lambda t)^2-t) M_t^{(\lambda)} \, d\mathbb{P}.$$
Putting $\lambda=0$, we see that $(B_t^2-t)_{t \geq 0}$ is a martingale. To get a martingale involving $B_t^4$ we need two differentiate two more times. Doing so yields
$$\int_F (B_s-\lambda s)((B_s-\lambda s)^2-3s) M_s^{(\lambda)} \, d\mathbb{P} = \int_F (B_t-\lambda t)((B_t-\lambda t)^2-3t) M_t^{(\lambda)} \, d\mathbb{P}$$
(which shows, for $\lambda=0$, that $(B_t^3-tB_t)_{t \geq 0}$ is a martingale) and
\begin{align*} &\int_F \left[((B_s-\lambda s)^2-s)((B_s-\lambda s)^2-3s)-2s(B_s-\lambda s)^2\right] M_s^{(\lambda)} \, d\mathbb{P} \\ &= \int_F\left[((B_t-\lambda t)^2-t)((B_t-\lambda t)^2-3t)-2t(B_t-\lambda t)^2 \right] M_t^{(\lambda)} \, d\mathbb{P},\end{align*}
which shows, for $\lambda=0$, that
$$(B_t^2-t)(B_t^2-3t)-2t B_t^2=B_t^4-6tB_t+3t^2$$
is a martingale.
Remark: There is a close connection to Hermite polynomials, see e.g. this question.