You mentioned Silver's theorem, which shows that if $\kappa$ is inaccessible in $V$, then there are no Kurepa trees after the Levy collapse making $\kappa=\omega_2$. Let me explain how to modify that argument so as also to make $2^{\omega_1}$ as large as desired.

Start in $V$, where $\kappa$ is inaccessible and the GCH holds. Consider the product forcing $P\times Q$, where $P=\text{Coll}(\omega_1,\lt\kappa)$ is the forcing that you used, and $Q=\text{Add}(\omega_1,\theta)$ is the forcing to add $\theta$ many Cohen subsets of $\omega_1$. Suppose that $V[G][H]$ is $V$-generic for this forcing. Note that both forcing notions are countably closed, and therefore the definition of $P$ and $Q$ is absolute to any of the extensions. In particular, the $Q$ forcing is countably closed in $V[G]$, and also $\kappa$-c.c., so cardinals $\kappa$ and above are all preserved.

By the result of Silver, we know that there are no Kurepa trees in $V[G]$, since this is the Levy collapse you mentioned.

Next, I claim that no $\omega_1$-tree $T$ in $V[G]$ becomes Kurepa in $V[G][H]$. This is because countably closed forcing cannot add any new branches to an $\omega_1$ tree in the ground model. If we had a name for a new branch $\dot b$, then we could decide this name in various incompatible ways, and get a level of $T$ that must have continuum many elements on it, contrary to $T$ being an $\omega_1$ tree.

Suppose that there is a Kurepa tree $T$ in $V[G][H]$, created by the $H$ forcing over $V[G]$. Since $T$ has size $\omega_1$ and the $Q$ forcing is $\omega_2$-c.c., it follows that $T$ exists in $V[G][H|A]$ for some subset $A\subset \theta$ of size $\omega_1$. We may rearrange the forcing and assume without loss that $T\in V[G][H|\omega_1]$ is added by the first $\omega_1$ many stages of the forcing. Indeed, since adding $\omega_1$ many subsets to $\omega_1$ is the same as adding just one, we may assume that $T\in V[G][H_0]$, where $H_0$ is the first subset to $\omega_1$ added by $H$. What is more, the rest of the $H$ forcing remains countably closed over this model, and therefore adds no new branches to $T$ that are not already in $V[G][H_0]$. Thus, all the branches of $T$ of $V[G][H_0]$. But now, the point is that doing the Levy collapse and then adding one subset of $\omega_1$ is isomorphic to just doing the Levy collapse. The later forcing is absorbed by the Levy collapse. (I can explain this if you like.) Thus, $V[G][H_0]=V[G^\ast]$ for some $V$-generic filter $G^\ast\subset P$. In particular, $T$ is not Kurepa there. So $T$ does not have $\omega_2$ many branches there, and hence does not have $\omega_2$ many branches in $V[G][H]$. So the model $V[G][H]$ has no Kurepa trees, yet $2^{\omega_1}\geq\theta$ there, as desired.

The answer is yes, and merely forcing over your model to add additional subsets of $\omega_1$ will pump up the value of $2^{\omega_1}$, while not creating Kurepa trees.

Specifically, let us start in $V$, where $\kappa$ is an inaccessible cardinal, and suppose also that the GCH holds. You mentioned the result of Silver, which shows that in the forcing extension obtained via the Levy collapse making $\kappa=\omega_2$, there are no Kurepa trees. I propose simply to add more subsets to $\omega_1$ over this model, and claim that still no Kurepa trees will be created.

To see this, consider the product forcing $P\times Q$, where $P=\text{Coll}(\omega_1,\lt\kappa)$ is the relevant Levy collapse, and $Q=\text{Add}(\omega_1,\theta)$ is the forcing to add $\theta$ many Cohen subsets of $\omega_1$. Suppose that $V[G][H]$ is $V$-generic for this forcing. Note that because the Levy collapse is countably closed, the forcing $Q$ is $\text{Add}(\omega_1,\theta)$ in both $V$ and in $V[G]$. In particular, the $Q$ forcing is countably closed in $V[G]$ and $\kappa$-c.c. there, so cardinals $\kappa$ and above are all preserved.

By the result of Silver, we know that there are no Kurepa trees in $V[G]$, since this is the Levy collapse you mentioned.

Next, I claim that no $\omega_1$-tree $T$ in $V[G]$ becomes Kurepa in $V[G][H]$. This is because countably closed forcing cannot add any new branches to an $\omega_1$ tree in the ground model. If we had a name for a new branch $\dot b$, then we could decide this name in various incompatible ways, and get a level of $T$ that must have continuum many elements on it, contrary to $T$ being an $\omega_1$ tree. (And this observation was critical to Silver's argument.)

Suppose that there is a Kurepa tree $T$ in $V[G][H]$, created by the $H$ forcing over $V[G]$. Since $T$ has size $\omega_1$ and the $Q$ forcing is $\omega_2$-c.c., it follows that $T$ exists in $V[G][H|A]$ for some subset $A\subset \theta$ of size $\omega_1$. We may rearrange the forcing and assume without loss that $T\in V[G][H|\omega_1]$ is added by the first $\omega_1$ many stages of the forcing. Indeed, since adding $\omega_1$ many subsets to $\omega_1$ is the same as adding just one, we may assume that $T\in V[G][H_0]$, where $H_0$ is the first subset to $\omega_1$ added by $H$. What is more, the rest of the $H$ forcing remains countably closed over this model, and therefore adds no new branches to $T$ that are not already in $V[G][H_0]$. Thus, all the branches of $T$ of $V[G][H]$ are in $V[G][H_0]$. But now, the point is that doing the Levy collapse and then adding one subset of $\omega_1$ is isomorphic to just doing the Levy collapse. The later forcing is absorbed by the Levy collapse. (I can explain this if you like.) Thus, $V[G][H_0]=V[G^\ast]$ for some $V$-generic filter $G^\ast\subset P$. In particular, $T$ is not Kurepa there. So $T$ does not have $\omega_2$ many branches there, and hence does not have $\omega_2$ many branches in $V[G][H]$. So the model $V[G][H]$ has no Kurepa trees, yet $2^{\omega_1}\geq\theta$ there, as desired.