If you make your requirement only for regular cardinals $\kappa$, then we can easily get an equiconsistency.

**Theorem.** The following theories are equiconsistent over ZFC:

- There are unboundedly many inaccessible cardinals.
- The continuum function $\kappa\mapsto 2^\kappa$ is injective and $2^\kappa$ is weakly inaccessible for any infinite regular cardinal $\kappa$.

Proof. Statement 2 implies that there are unboundedly many (strongly) inaccessible cardinals in $L$, and so statement 1 holds in $L$. Conversely, assume statement 1 holds. By forcing, we may assume without loss that the GCH also holds in $V$. Let $\kappa_\alpha$ be the $\alpha^{th}$ infinite regular cardinal and let $\delta_\alpha$ be the $\alpha^{th}$ inaccessible cardinal. Define $$E(\kappa_\alpha)=\delta_{\alpha+1}.$$ This function is increasing and has $\text{cof}(E(\kappa))>\kappa$ for every regular cardinal $\kappa$. Thus, it satisfies the hypotheses of Easton's theorem, and so there is a cardinal and cofinality-preserving forcing extension $V[G]$ in which $2^\kappa=E(\kappa)$ for every regular cardinal $\kappa$. Since the $\delta_\alpha$ all remain weakly inaccessible in the extension, we have that $2^\kappa$ is weakly inaccessible for every regular cardinal $\kappa$. Furthermore, the continuum function in $V[G]$ is strictly increasing on successor cardinals and hence is injective. QED

Meanwhile, if you really want $2^\kappa$ to be weakly inaccessible even for singular $\kappa$, then the hypothesis becomes inconsistent with the singular cardinals hypothesis, and hence inconsistent with the existence of supercompact or strongly compact cardinals.

**Theorem.** If $2^\kappa$ is weakly inaccessible for all infinite cardinals $\kappa$, then the SCH fails at every singular strong limit. In particular, this hypothesis is inconsistent with the existence of supercompact or strongly compact cardinals.

Proof. If $\kappa$ is a singular strong limit, then under the SCH, it must be that $2^\kappa=\kappa^+$, which is definitely not weakly inaccessible. QED

The failure of SCH already has a fairly high large cardinal strength, and so we get lower bounds for the consistency strength of your hypothesis (when applied to all cardinals including singular cardinals). Perhaps the excellent inner model theory experts we have here on MO can state more precise claims about this.