From the exchanges in the comments, one reaches the conclusion that $M_t=K_{A(t)}$ with $ K_t=\int_0^t\mathrm e^s\mathrm dB_s,\qquad A(t)=\log\sqrt{1+2t}. $ Since $(K_t)_{t\geqslant0}$ is a martingale and the (deterministic) function $A$ is nondecreasing, this is enough to show that $(M_t)_{t\geqslant0}$ is a martingale. Furthermore, for every $0\leqslant s\leqslant t$, by Itô's isometry, the identity $ (M_t-M_s)^2=\left(\int_{A(s)}^{A(t)}\mathrm e^u\mathrm dB_u\right)^2 $ implies that $ \mathbb E((M_t-M_s)^2)=\mathbb E\left(\int_{A(s)}^{A(t)}\mathrm e^{2u}\mathrm du\right)=\frac{\mathrm e^{2A(t)}-\mathrm e^{2A(s)}}2. $ The function $A$ is tuned such that $\mathbb E((M_t-M_s)^2)=t-s$ hence $(M_t)_{t\geqslant0}$ is a Brownian motion. More generally, for every nonzero function $a$, $(M^a_t)_{t\geqslant0}$ is a Brownian motion, where $ M^a_t=\int_0^{A^{-1}(t)}a(s)\mathrm dB_s,\qquad A(t)=\int_0^{t}a(s)^2\mathrm ds. $ The two expectatons to be computed are direct since one can replace $(M_t)_{t\geqslant0}$ by $(B_t)_{t\geqslant0}$ without changing the result. For example, $ X_t=\int_0^tB_t^6\mathrm dB_t $ is an odd functional of $(B_s)_{0\leqslant s\leqslant t}$, hence $\mathbb E(X_t)=0$. Likewise, $ Y_t=\int_0^tB_s\mathrm dB_s $ is a centered gaussian random variable hence $\mathbb E(Y_t^3)=0$.
On time-changed Brownian motions, see this.
Edit: Since $A$ is invertible and continuous, the sigma-algebras $\mathcal F_t^B=\sigma(B_s;s\leqslant t)$ and $\mathcal F_t^M=\sigma(M_s;s\leqslant t)$ are such that $\mathcal F^M_t=\mathcal F^B_{A(t)}$. For every $s\leqslant t$, $M_t=M_s+\Delta$ where $ \Delta=\int_{A(s)}^{A(t)}\mathrm e^u\mathrm dB_u. $ The increments of $(B_u)_{u\geqslant A(s)}$ are independent of $\mathcal F^B_{A(s)}$ hence $\mathbb E(\Delta\mid\mathcal F^B_{A(s)})=0$, which shows that $\mathbb E(M_t\mid\mathcal F^M_s)=M_s$. Hence $(M_t)_{t\geqslant0}$ is a martingale.