In discrete-time, it is possible to construct a martingale which is in $L^p$ but not $L^q$ for $p<q$. To see this, simply let $(\Omega, \mathcal{F}, \mathcal{F}_n, \mathbb{P})$ be a probability space for which $(\epsilon_n)$ is an iid sequence of centred random variables in $L^p \setminus L^q$. Now construct the random walk $X_0=0$,
$$X_n = \sum_{k=1}^n \epsilon_n$$
This is now a martingale with the desired properties.
Is this possible to construct a continuous martingale with the same property? Namely, can one have a (true, non-trivial) continuous martingale $X_t$, with $X_0 = 0$, such that $X_t$ has finite $p$-th moment, but all higher moments are infinite? By non-trivial I mean non-constant.
Fix $ c \in \left(0,\frac 12\right)$ let $M_t := \mathbb{E}[e^{c B_1^2}|\mathcal F_t]$ for $t \le 1$, where $B$ is a Brownian motion and $\mathcal F$ is the filtration generated by $B$. Since $M$ is a martingale with respect to the Brownian filtration, it has a continuous modification which we consider from here on out. Note that $M_1 = e^{c B_1^2}$ so $\mathbb{E}[M_1^p] = \mathbb{E}[e^{pc B_1^2}]$ is finite if $pc < \frac 12$ and infinite if $pc \ge \frac 12$, so at least $M_1$ satisfies what you want. For each $t > 0$ we'll find that there exist $q,p > 0$ such that $M_t \in L^p \setminus L^q$.
For $t < 1$, we can directly compute $M_t$ using the independent increments property:
\begin{align*} \mathbb{E}[e^{c B_1^2}|\mathcal F_t] &= \mathbb{E}[e^{c (B_1-B_t)^2 + 2c B_t (B_1-B_t) + cB_t^2}|\mathcal F_t] \\ &= e^{cB_t^2} \mathbb{E}[e^{c (1-t)Z^2 + 2c B_t \sqrt{1-t}Z}|\mathcal F_t] \end{align*} where $Z \sim N(0,1)$ is independent of $\mathcal F_t$. By the independence lemma, we have \begin{align*} \mathbb{E}[e^{c (1-t)Z^2 + 2c B_t \sqrt{1-t}Z}|\mathcal F_t] &= \frac 1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{c (1-t)z^2 + (2c B_t \sqrt{1-t})z-\frac 12 z^2}dz \end{align*} and by completing the square (please double check these computations) we have \begin{align*} c (1-t)z^2 + (2c B_t \sqrt{1-t})z-\frac 12 z^2 &= -\frac 12 a(z-k)^2 + h \end{align*} where \begin{align*} a &= -2\left(c(1-t)-\frac 12\right) = 1-2c(1-t) > 1-2c > 0 \\ k &= 2c\frac{\sqrt{1-t}B_t}{a} \\ h &= \frac 12 ak^2. \end{align*}
Let $\sigma = \frac{1}{\sqrt{a}}$ so \begin{align*} \frac 1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{c (1-t)z^2 + (2c B_t \sqrt{1-t})z-\frac 12 z^2}dz &= \frac 1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac 12 a(z-k)^2+h}dz \\ &= \sigma e^{h}\frac 1{\sigma \sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac 1{2\sigma^2} (z-k)^2}dz \\ &= \sigma e^{h} = \sigma \exp \left(\left(\frac{c^2}{2a}(1-t)\right)B_t^2 \right) \end{align*} and therefore \begin{align*}M_t = e^{cB_t^2} \mathbb{E}[e^{c (1-t)Z^2 + 2c B_t \sqrt{1-t}Z}|\mathcal F_t] &= \sigma \exp \left(\left(\frac{c^2}{2a}(1-t) + c\right)B_t^2 \right) \\ & =\sigma \exp \left(\left(\frac{(1-t)c^2}{2-4c(1-t)} + c\right)B_t^2 \right) .\end{align*} Since $a \ge 1-2c$ for all $t$, we have $\sigma \le \frac{1}{\sqrt{1-2c}}$ for all $t$ and therefore we conclude $M_t \in L^p$ if and only if $$\left(\frac{(1-t)c^2}{2-4c(1-t)} + c\right)p < \frac{1}{2t}.$$ Since $\frac{1-t}{2-4c(1-t)} > 0$ and $c > 0$, for all $t > 0$ we can find some $p$ where the inequality holds and some $p$ where it doesn't. Unfortunately we don't have a single $p$ and $q$ where $M_t \in L^p\setminus L^q$ for all $t > 0$, but we do have that for all $t_0 > 0$ there exists $p_0 \in (1,\infty)$ such that $M_t \not \in L^p$ for all $t \ge t_0$ and $p \ge p_0$. Note that if $M_1 \in L^p$ then $M_t \in L^p$ for all $t < 1$, so we also have that $M_t \in L^p$ for all $p < \frac{1}{2c}$ and $t \in [0,1]$.
We can replace $M_t$ by $M_t - M_0 = M_t - \mathbb{E}[e^{cB_1^2}]$ to get $M_0=0$ if you would like.