Notice first that any perfect cube is congruent to either $0$, $1$ or $6$ modulo $7$ and it is congruent to either $0$, $1$, $5$, $8$, or $12$ modulo $13$. Because $7$ and $13$ are relatively prime numbers, by the Chinese Remainder theorem, a perfect cube is congruent to precisely one of $15 = 3 \times 5$ different residues modulo $91 = 7 \times 13$. Therefore, by the Pigeonhole principle, we can always find among any $16$ perfect cubes, two cubes which are congruent modulo $91$.