Cardinality of power sets decides all of cardinal arithmetic?

The answer is no assuming the existence of a model of $\mathsf{ZFC+GCH}$ having a supercompact cardinal.

First, using Silver's forcing, there is a generic extension $V[K]$ where $\kappa$ is still measurable but $2^{\kappa}=\kappa^{++}$. Then we can use Prikry's forcing to obtain a generic extension $V[K][H]$ of $V[K]$ such that all bounded subsets of $\kappa$ are in $V[K]$, all cardinals are preserved, $\kappa$ is still a strong limit and $\operatorname{cf}\kappa=\omega$. Let $G=K\ast H$.

As we assumed $V\models\mathsf{GCH}$, it's not hard to prove that in $V[G]$ we have $2^\lambda=\lambda^+$ for all $\lambda>\kappa$; since the poset yielding $V[K]$ has size $\kappa^{++}$ in $V$ and the poset giving the extension $V[K][H]$ has size $\kappa^{++}$ in $V[K]$.

Now let us work in $V[G]$. We have $$\kappa^{\aleph_0}=\kappa^{\operatorname{cf}\kappa}=2^{\kappa}=\kappa^{++},$$ and $$(\kappa^{+3})^{\omega_1}=(2^{\kappa^{++}})^{\omega_1}=2^{\kappa^{++}}=\kappa^{+3},$$ thus we can force with $Add(\omega_1,\kappa^{+3})$ to obtain a generic extension $V[G][H']$ where $2^{\omega_1}=\kappa^{+3}$, and $2^{\lambda}=\kappa^{+3}$ for all $\omega_1\leq\lambda\leq \kappa^{++}$. In $V[K]$, $\kappa$ is measurable, thus in there $\kappa=\aleph_\kappa$, so as cardinals are preserved in $V[G]$ we get that $\kappa=\aleph_\kappa$ is also true in $V[G][H']$.

Let $\beta_0$ be such that $2^{\aleph_0}=\aleph_{\beta_0}$ in $V[G]$. Then $\beta_0<\kappa$; as all bounded subsets of $\kappa$ in $V[G]$ are in $V[K]$. We also have $2^{\aleph_0}=\aleph_{\beta_0}$ in $V[G][H']$.

Thus if we consider the following function $$F(\alpha)=\begin{cases} \beta_0 & \text{if}&\alpha=0 \\\kappa+3 & \text{if}& 1\leq\alpha\leq\kappa+2\\\alpha+1 &\text{if}&\alpha\geq\kappa+3 \end{cases},$$ it follows that for all ordinals $\alpha$, $$V[G][H']\models 2^{\aleph_\alpha}=\aleph_{F(\alpha)},$$ and as the poset we used in $V[G]$ is $<\omega_1$-closed there, we get that $\kappa^{\aleph_0}=\kappa^{++}$ in this extension too.

Now, let $\mathbb P\in L$ be a poset, cardinal preserving, such that if $K'$ is $L$-generic over $\mathbb P$, we have for all ordinals $\alpha$, $$L[K']\models 2^{\aleph_\alpha}=\aleph_{F(\alpha)}.$$ In $L$, $\kappa$ is inaccesible, and thus in this model $\aleph_\kappa=\kappa$, so as $\mathbb P$ preserves cardinals we get that $\aleph_\kappa=\kappa$ in $L[K']$ too.

Let us work in $L[K']$. The singular cardinals hypothesis is true; since $0^\sharp$ does not exist, so we get that as $2^{\aleph_0}<\kappa$ and $\kappa$ is regular, $\kappa^{\aleph_0}=\kappa$.

Therefore we have that in both models $V[G][H']$ and $L[K']$, $2^{\aleph_\alpha}=\aleph_{F(\alpha)}$ for all ordinals $\alpha$, but $$V[G][H']\models \aleph_\kappa^{\aleph_0}=\aleph_{\kappa+2}\text{ and }L[K']\models \aleph_\kappa^{\aleph_0}=\aleph_\kappa.$$

Note: This argument should go through with no problem using just a measurable cardinal $\kappa$ of Mitchell order $\kappa^{++}$, working in Mitchell's model for such $\kappa$, using Gitik and Woodin's forcing. However as I'm not that familiar with this method, I used Silver's instead.