This is just a translation of @WillBrian's answer into a statement about dense sets.

For each $\delta \in \mathsf{On}$, fix $\dot{R}_\delta \in V^{\mathbb{P}}$ of minimal rank, such that $1 \Vdash_\mathbb{P} \dot{R}_\delta = \{ x : \mathsf{rank}(x) < \check{\delta}\}$. Then, because of how the rank function works, we always have that the least $\delta \in \mathsf{On}$, such that $1 \Vdash_{\mathbb{P}} \dot{R}_\delta \neq \check{V}_\delta$ is a successor; so the thing to do is characterize when $1 \Vdash_{\mathbb{P}} \dot{R}_{\gamma + 1} = \check{V}_{\gamma + 1}$. To this end,

**Theorem:** For every $\delta \in \mathsf{On}$ such that $1 \Vdash_{\mathbb{P}} \dot{R}_\delta = \check{V}_\delta$, the following are equivalent,

$1 \Vdash_{\mathbb{P}} \dot{R}_{\delta+1} = \check{V}_{\delta+1}$,

For every $p \in \mathbb{P}$ and every $\{ \mathcal{B}_x: x \in V_{\delta} \} \subset \mathcal{P}(\mathbb{P})$, there is some $p_0 \le p$, such that, for any $x \in V_{\delta}$, either
$$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}\text{, or }\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\},$$
is dense below $p_0$.

**Proof:** Fix $\delta \in \mathsf{On}$, and assume that $1 \Vdash_{\mathbb{P}} \dot{R}_{\delta+1} = \check{V}_{\delta+1}$. Then, for every $p\in \mathbb{P}$ and $\dot{g} \in V^{\mathbb{P}}$ such that $p \Vdash_{\mathbb{P}} \dot{g}:\check{V}_\delta \rightarrow \check{2}$, the set $\{ q \le p : (\exists f: V_\delta\rightarrow 2)(q \Vdash_{\mathbb{P}} \dot{g} = \check{f} ) \}$ is dense. So let $\{ \mathcal{B}_x : x \in V_\delta\} \subset \mathcal{P}(\mathbb{P})$ be arbitrary, and define $\dot{g},\sigma_x, \tau_x \in V^{\mathbb{P}}$ ($x\in V_\delta$) as follows $\dot{g} = \{ \langle\sigma_x, 1\rangle: x\in V_\delta \}$, $\sigma_x = \mathsf{pair}_\mathbb{P}(\check{x}, \tau_x)$, and $\tau_x = \{ \langle \emptyset, p \rangle: p \in \mathcal{B}_x \}$. Then, $1 \Vdash_{\mathbb{P}} \dot{g}: \check{V}_\delta \rightarrow \check{2}$ and so by assumption the set $\{ q \in \mathbb{P}: (\exists f: V_\delta\rightarrow 2)(q \Vdash_{\mathbb{P}} \dot{g} = \check{f} )\}$ is dense open, hence given $p \in \mathbb{P}$ we can find $p_0 \le p$ and $f:V_\delta \rightarrow 2$ such that, for every $x \in V_\delta$, $p_0 \Vdash \dot{g}(\check{x}) = \check{f}(\check{x})$ or equivalently, $ p_0 \Vdash_{\mathbb{P}} "\tau_x = \check{f}(\check{x})" \iff$ $f(x) = 1$ and $\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}$ is dense below $p_0$, or $f(x)=0$ and $\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\}$ is dense below $p_0$.

For the converse, assume (2), then fix $p\in\mathbb{P}$ and $\dot{a}\in V^{\mathbb{P}}$ such that $p \Vdash_\mathbb{P} \dot{a} \in \dot{R}_{\delta+1}$. Then, for $x\in V_\delta$ defining $\mathcal{B}_x = \{ q \in \mathbb{P}: q\Vdash_{\mathbb{P}} \check{x} \in \dot{a} \}$, we can find $p_0 \le p$ (by 2) such that for every $x$, either
$$\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}\text{, or }\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\},$$
is dense below $p_0$. As such, letting $$A = \{ x \in V_\delta : \{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\} \text{ is dense below } p_0 \}$$
we have $p_0 \Vdash \dot{a} = \check{A} \in \check{V}_{\delta+1}$. $\square$

**Remark:** If in (2), the requirement "either $A^x_1=\{q_0 \in \mathbb{P}: (\exists s \in \mathcal{B}_x)( q_0 \le s)\}$, or $A^x_0=\{q_0 \in \mathbb{P}: (\forall s \in \mathcal{B}_x)( q_0 \perp s)\}$ is dense below $p_0$", is changed to "$p_0$ is an element of $A^x_0$ or $A^x_1$", then the new statement is equivalent to $\mathbb{P}$ being $\vert V_\delta \vert$-baire.