Suppose $2 \nmid n$. Then all prime divisors of $n$ are odd. But if $p | n$ is an odd prime, $p^2 - 1 | n$ is even, contradiction. Therefore, $2 | n$. Then $2^2 - 1 = 3 | n$ as well. Next, $3^2 - 1 = 8 | n$ so $24 | n$. It is now simple to check multiples of $24$ near $2012$. The first few, descending, are $1992, 1968, 1944$. $1992$ cannot work since $83 | 1992$ but $83^2 - 1 = 6888 \nmid 1992$. $1968$ cannot work either, since $41 | 1992$ but $41^2 - 1 = 1680 \nmid 1992$. $1994$ works; it factors as $2^3*3^4$ where $2^2 - 1 = 3 | 1994$ and $3^2 - 1 = 8 | 1994$ as desired.

Just a little point: The $3rd$ number is $1944$ and it's $2^3.3^5.$

@Hyperbolictangent, well done. Your solution is similar with mine.