In Dobrow's Introduction to Stochastic Processes, we are given the following statement:
Here is one way to tell if a right stochastic matrix (i.e., a square matrix with non-negative entries whose rows sum to 1) is not regular. If for some power $n$, all the $0$s in $P^n$ appear in the same location as all the $0$s in $P^{n+1}$, then they will appear in the same locations for all higher powers, and the matrix is not regular.
However, I am unable to find a proof anywhere. My guess is that it is related to the $0$ entries representing the set of inaccessible pairs of states after $n$ and $n+1$ steps, but I have not been able to go any further than that.
Fix $d$ as a positive integer that represents the dimension. For $i \in \{1, ..., d\}$, let $e_i\in\mathbb{R}^d$ be the unit vector that is 1 in entry $i$ and 0 elsewhere. Note that if $A=(A_{ij})$ is a $d\times d$ matrix then $$ e_i^{\top}Ae_j=A_{ij} \quad \forall i,j \in \{1, ..., d\}$$ Further, if all entries of $A$ are nonnegative, then for all positive integers $n$, the matrix $A^n$ has nonnegative entries.
Claim: Suppose $P$ is a $d\times d$ matrix with nonnegative entries. Suppose there is a positive integer $n$ such that $$\left(e_i^{\top} P^ne_j=0 \iff e_i^{\top}P^{n+1}e_j=0\right)\quad\forall i,j\in\{1,...d\}$$ Then
$$\left(e_i^{\top} P^{n+1}e_j=0 \iff e_i^{\top}P^{n+2}e_j=0\right)\quad\forall i,j\in\{1,...d\}$$
Proof: Define the nonnegative matrices $Q=P^n$ and $R=P^{n+1}$. We are told that $$\left(Q_{ij}=0\iff R_{ij}=0\right) \quad \forall i,j \in \{1, ..., d\} \quad (Eq. 1)$$
(Proof for direction $\implies$): Fix $(a,b) \in \{1, ..., d\}^2$. Suppose $R_{ab}=0$. We want to show $e_a^{\top}PRe_b=0$. Since $R=PQ$ we have $$ 0=R_{ab}=\sum_{i=1}^dP_{ai}Q_{ib}$$ and since $P_{ai}Q_{ib}\geq 0$ for all $i$, we know $$ \left(P_{ai}>0\implies Q_{ib}=0\right) \quad \forall i\in\{1, ..., d\}$$ and hence by (Eq. 1) $$ \left(P_{ai}>0\implies R_{ib}=0\right) \quad \forall i\in\{1, ..., d\} \quad (Eq. 2)$$ Now we have \begin{align} e_a^{\top}PRe_b &= \sum_{i=1}^dP_{ai}R_{ib}\\ &=\sum_{i:P_{ai}>0}P_{ai}R_{ib}\\ &=0 \end{align} where the last equality holds by (Eq. 2). $\Box$
(Proof for direction $\impliedby$): Fix $(a,b) \in \{1, ..., d\}^2$. Suppose $e_a^{\top}PRe_b=0$. We want to show $R_{ab}=0$. We know $$ 0 = e_a^{\top}PRe_b=\sum_{i=1}^dP_{ai}R_{ib}$$ and since $P_{ai}R_{ib}\geq 0$ for all $i$ we know $$ \left(P_{ai}>0\implies R_{ib}=0 \implies Q_{ib}=0\right) \quad \forall i \in \{1 ,... ,d\} \quad (Eq. 3)$$ where we have again used (Eq. 1). Then, since $R=PQ$, \begin{align} R_{ab} &= \sum_{i=1}^dP_{ai}Q_{ib}\\ &=\sum_{i:P_{ai}>0}P_{ai}Q_{ib}\\ &=0 \end{align} where the last equality uses (Eq. 3). $\Box$
We can recursively apply the Claim to show that if the hypothesis of the Claim holds then for all positive integers $k$: $$\left(e_i^{\top}P^ne_j=0 \iff e_i^{\top}P^{n+k}e_j=0\right) \quad \forall i,j\in\{1, ..., d\}$$