No, you cannot prove this because it is not true. For example, consider $M$ to be the product of two oriented, complete Riemannian surfaces $M=\Sigma_1\times\Sigma_2$ where $g$ is the product metric from the metric $g_i$ on $\Sigma_i$. Let $J_i$ be the $g_i$-compatible complex structure on $\Sigma_i$ that induces the given orientation on $\Sigma_i$ and let $I = J_1 + J_2$ and $J = J_1 - J_2$. Then $I$ and $J$ are $g$-parallel and $g$-compatible almost complex structures on $M$. However, except in the very special situation that each $g_i$ is flat, $g$ is not hyperKähler since it is not Ricci-flat.
Meanwhile, if you have a connected Riemannian $4$-manifold $(M^4,g)$ that supports two $g$-compatible, $g$-parallel almost complex structures $I$ and $J$ that induce the same orientation on $M$ and do not satisfy $I = \pm J$ , then, yes, $(M^4,g)$ supports a $g$-compatible hyperKähler structure. The reason is simple: Under these hypotheses, with $M$ oriented compatibly with the common orientation induced by $I$ and $J$, then $I$ and $J$ correspond, respectively, to self-dual $2$-forms $\omega_I$ and $\omega_J$ that are $g$-parallel and linearly independent everywhere. This implies that the induced connection on the rank-3 bundle of self-dual $2$-forms on $M$ is flat (since the connection is metric and the bundle has a flat subbundle of rank-$2$). This is equivalent to $(M^4,g)$ supporting a $g$-compatible hyperKähler structure.