Auxiliary result related to the exponential martingale inequality Let $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space and $(\mathcal F_t)_{t\ge0}$ be a complete filtration on $(\Omega,\mathcal A,\operatorname P)$.
Let $(M_t)_{t\ge0}$ be a local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$. By the Itō formula, $$N^\sigma:=e^{-\frac{\sigma^2}2[M]+\sigma M}=N_0+\sigma N\cdot M\tag1$$ is a local $\mathcal F$-martingale for all $\sigma\in\mathbb R$.

Now assume $(M_t)_{t\ge0}$ is a continuous $\mathcal F$-martingale and $$\operatorname E\left[e^{\lambda[M]_t}\right]<\infty\;\;\;\text{for all }t>0\text{ and }\lambda>0\tag2.$$ Are we able to conclude that $N^\sigma$ is a $\mathcal F$-martingale for all $\sigma\in\mathbb R$?

Let me stress one subtlety, which might be involved here: If $M$ is square-integrable, we know that $N\cdot M$ is a square-integrable $\mathcal F$-martingale if $$\operatorname E\left[\int_0^t|N_s|^2\:{\rm d}[M]_s\right]<\infty\;\;\;\text{for all }t>0\tag3.$$ I'm not sure whether the square-integrability of $M$ is really necessary for the martingale conclusion to hold (it is surely necessary to obtain the square-integrability of $N\cdot M$, but in the context of this question we are not interested in this integrability conclusion).
So, maybe we need to assume that $M$ is square-integrable. Ignoring this for a moment, we clearly can use that $e^x\le1$ for all $x\le0$ and hence \begin{equation}\begin{split}\operatorname E\left[\int_0^t|N_s|^2\:{\rm d}[M]_s\right]&=\operatorname E\left[\int_0^te^{-\sigma^2[M]_s+2\sigma M_s}\:{\rm d}[M]_s\right]\\&\le\operatorname E\left[\int_0^te^{2\sigma M_s}\:{\rm d}[M]_s\right]\end{split}\tag4\end{equation} for all $t>0$.

Does the assumption $(2)$ somehow imply that $(4)$ is finite for all $t>0$?

 A: The condition in $(2)$ is enough to guarantee $N^\sigma$ is a martingale by Novikov's condition.  Let $\mathcal E(M)_t := e^{M_t - \frac{1}{2} [M]_t}$ and notice that $N^\sigma = \mathcal E(\sigma M)$.  Then since $\mathbb{E}[e^{\frac 12 [\sigma M]_t}] = \mathbb{E}[e^{\frac{\sigma^2}2 [M]_t}]  < \infty$, Novikov's condition gives $N^\sigma$ is a martingale.
We don't need to assume $M_t$ is square integrable because $(2)$ actually implies that $M_t$ has finite moments of all orders.  From a Taylor expansion, $\mathbb{E}[[M]_t^p] \le c_p \mathbb{E}[e^{[M]_t}] < \infty$ so the BDG inequality gives $\mathbb{E}[\sup_{s \le t} |M_s|^p] \le C_p \mathbb{E}[[M]_t^{p/2}] < \infty$ for all $p > 0$.
To get $N^\sigma_t$ is square integrable, we can apply Holder's inequality: \begin{align*}
\mathbb{E}[(N^\sigma_t)^2] &= \mathbb{E}[e^{2\sigma M_t - \sigma^2 [M]_t}] \\
&= \mathbb{E}[(e^{4\sigma M_t - 8 \sigma^2 [M]_t})^{1/2}e^{4 \sigma^2 [M]_t}] \\
&= \mathbb{E}[(N^{4 \sigma}_t)^{1/2}e^{4 \sigma^2 [M]_t}] \\
&\le \mathbb{E}[N^{4 \sigma}_t]^{1/2} \mathbb{E}[e^{8 \sigma [M]_t}]^{1/2} < \infty.
\end{align*}
