Statement I gives us the equation $xy^2 = 116$, which can be factored into $2^2 \times 29$. Testing factors of $116$ shows $x$ can be $2$ or $29$, making $x$ potentially not prime if $x=2$ and thus indecisive. Statement II, $xz = 261$, factors as $3 \times 87$ or $9 \times 29$, and since $x > 1$ and must result in an integer greater than $1$, primes $3$ and $29$ are possible. Therefore, \textbf{only Statement II conclusively gives $x$ as prime}.