The answer depends on your field $k$. For example if $k$ is the $p$-adic field $\mathbb{Q}_p$, $p\neq 2$, it is known that the only anisotropic form of dimension $4$ over $k$ is isomorphic to the norm form of the unique quaternion algebra over $k$ which is of course in $I^2$ since it is a Pfister form. By contrast you can construct many fields $k$ with your desired property, for instance consider the field $k=\mathbb{Q}(x,y,z,t)$ and the form $\varphi=\langle x,y,z,t\rangle\in I\backslash I^2$.

The answer depends on your field $k$. For example if $k$ is the $p$-adic field $\mathbb{Q}_p$, $p\neq 2$, it is known that the only anisotropic form of dimension $4$ over $k$ is isomorphic to the norm form of the unique quaternion algebra over $k$ which is of course in $I^2$ since it is a Pfister form. By contrast you can construct many fields $k$ with your desired property, for instance consider the field $k=\mathbb{Q}(x,y,z,t)$ and the form $\varphi=\langle x,y,z,t\rangle\in I\backslash I^2$.

Added: regarding your question in the comments about $I^3$:

By a theorem of Merkurjev, the Hasse-Witt invariant induces an isomorphism between $I^2/I^3$ and the two torsion part of the Brauer group $Br_2(k)$. Hence it suffices to give an example of a field $k$ and an anisotropic form $\varphi\in I^2$ of dimension $8$ such that the Hasse-Witt invariant of $\varphi$ is nontrivial. For this consider an anisotropic Albert form, e.g., the form $\alpha=\langle x,y,-xy,-z,-t,zt\rangle$ over the transcendental field $\mathbb{Q}(x,y,z,t)$. Now consider the transcendental field $k=\mathbb{Q}(x,y,z,t,u,v)$ and the eight dimensional anisotropic form $\varphi=-uv\cdot\alpha\perp\langle u,v\rangle\in I^2$. The Hasse-Witt invariant of $\varphi$ is isomorphic to the Clifford algebra $Cl(\varphi)\simeq C(\langle u, v\rangle)\otimes Cl(\alpha)$ which is not split.