I added the "probability" tag (although it doesn't have to do anything with the solution) as Markov chain convergence was the motivation. A convergence criterion is "no zeroes in some power of M", where M is the transition matrix.
Let's drop the condition that the rows sum to 1 and just insist that all elements of M are zero or positive. (Uhm, can the number of zeroes increase when going from a lower to a higher power of M?) Assume there is one finite m such that the m-th power of M (and thus all higher ones) have no zeroes. How is the maximum attainable m related to the size of M? Using
"M(i,i+1)=M(1,1)=M(1,n)=1,else M(i,j)=0"
I got m=6,12,18 for n=4,7,10. So m=2n-2?

For a matrix $M\in\mathbb{R}^{n\times n}_{\geq 0}$ with nonnegative entries, we define $m$ as the smallest positive integer such that all the entries of $M^m$ are strictly positive (if there is one).

What is, as a function of $n$, the maximum possible (finite) value of $m$ over all possible choices of $M$?

My attempt: By taking the matrix $M(i,i+1)=M(1,1)=M(1,n)=1$ else $M(i,j)=0$ I got $m=6,12,18$ for $n=4,7,10$. So it seems that $m=2n-2$?

I added the "probability" tag because Markov chain convergence was the motivation.