Edit: this answer only makes sense if the $X$ is different; see Robert Israel's answer instead.
Given your assumption $\lVert x \rVert_1 \leq M \lVert x \rVert_2$ for some $M > 0$, we can only say that the identity map $I: (X, \lVert \cdot \rVert_2) \to (X, \lVert \cdot \rVert_1)$ is continuous.
To see this, note that if $\lVert x \rVert_2 < \epsilon / M$, then $\lVert I(x) \rVert_1 = \lVert x \rVert_1 \leq M \lVert x \rVert_2 < \epsilon$.
The fact that $X$ associated with either metric being a Banach space doesn't really matter here; this is a more general result from metric/topological spaces. We see that the topology induced by the norm $\lVert \cdot \rVert_1$ is finer than the one induced by the norm $\lVert \cdot \rVert_2$.
If the two norms were equivalent, i.e. $m \lVert x \rVert_2 \leq \lVert x \rVert_1 \leq M \lVert x \rVert_2$, then their topologies would be the same, and so the identity map would be continuous.
For a counterexample to the original direction, we can consider the Banach spaces $\ell^1$ and $\ell^\infty$ (where $\ell^1 \subseteq \ell^\infty$ or equivalently $\Vert \cdot \rVert_\infty \leq \lVert \cdot \rVert_1$). Consider the identity map $I: \ell^\infty \to \ell^1$, and the element $x = (1, 1, 1, \dots)$ which has $\Vert x \rVert_\infty = 1$ but $\lVert x \rVert_1 = \infty$.