This is actually meant to be a comment, not an answer, but I'm new here and I don't have enough reputation to post a comment yet...sorry!
I just wanted to note that Dickson's conjecture would imply infinitely many consecutive numbers with the same prime signature.
For example, Dickson's conjecture would say that there are infinitely many k such that 4k+1 and 9k+2 are both prime. For each such k, 4(9k+2) and 9(4k+1) would be consecutive numbers with the prime signature (1,2). One might expect there to be roughly $\frac{3N}{\log 4N \log 9N}$ values of k between 1 and N such that 4k+1 and 9k+2 are both prime.
This additional comment is directed toward Davidac897's comment about having more than one prime factor with exponent greater than 1, which TonyK already pointed out that Tom Sirgedas' program has already found examples of.
Dickson's conjecture also would imply infinitely many such examples where more than one prime has exponent greater than 1.
For example, say we want prime signature (1,2,2). Let $a = 2^27^2$ and $b = 3^25^2$. We seek solutions in primes $p$ and $q$ to $ap + 1 = bq$. If $p = bk + 194$, then $q = ak + 169$. Dickson's conjecture would say there are infinitely many $k$ such that $bk+194$ and $ak + 169$ are both prime. (The first consecutive pair using this method is $2463524 = 2^27^212569$ and $2463525 = 3^25^210949$.)
In a similar vein, you can use Dickson's conjecture to force any prime signature you wish provided that at least one of the exponents is 1.