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Heuristically this should be the case. For any prime p greater than 5, consider the set of numbers of the form $2^a3^b5^c p \pm 1$. The "probability" that one of of these is prime should be about $$\frac{1}{\log (2^a 3^b 5^c)} = \frac{1}{a \log 2 + b \log 3 + c \log 5} .$$ So the probability that both $2^a3^b5^c p + 1$ and $2^a3^b5^c p - 1$ are prime should be about $$\frac{1}{\left(a \log 2 + b \log 3 + c \log 5\right)^2}.$$ Now, note that the series $$\sum_{a,b,c} \frac{1}{\left(a \log 2 + b \log 3 + c \log 5\right)^2}$$ diverges, so we should expect for a given $p$ there should be such an $a$,$b$ and $c$. So after noting the twin prime pairs $(3,5)$, $(5,7)$ and $(29,31)$ it seems like we should expect a much stronger statement. For any prime $p$, there should be a positive integer $n$ such that $p$ is the largest prime divisor of $n$ and $n+1$ and $n-1$ are both prime.