user2969 asked for the full answer so I'll post it here.
The first step is to take the gcd of the two numbers. This can be done with the Euclidean algorithm and is fast: the naive implementation takes quadratic time, which is good enough. Call this number g.
The next step is to factor g. This is the hardest step if g is large. Basic approach: divide g by the primes up to some fixed limit, say 10^5, or until what remains is less than the square of the current prime. At this point test what remains for primality (first with a Miller-Rabin test then, if desired, with the test of your choice, APR-CL, ECPP, etc.). If the number is composite, use Pollard's rho and/or Shanks' SQUFOF. (Whenever you find a factor, check it and its cofactor for primality and return to this step if composite.) If those fail to find a factor, use ECM to search for a small factor. At an appropriate crossover point you may wish to use MPQS/SIQS to factor the number instead. If the number is larger, increase the bounds on ECM. If no factor is found to ~40 digits, consider using GNFS to factor what remains. This will take at least a day, depending on the size of the remaining number.
Once you have the factorization it's easy to count the number of divisors: start an accumulator at 1, and for each prime multiply the accumulator by one more than the exponent of that prime.
Note that for this problem you may be able to short-circuit the factorization for numbers of an appropriate size by trial-dividing up to the cube root of the number and showing that what remains is composite and not a square. In this case the number of divisors is the number of divisors of the factored part times 4, since the unfactored part is a squarefree semiprime.
Here are some scripts to calculate the value.
common(m,n) = numdiv(gcd(m,n))
Common[m_,n_] := DivisorSigma[0,GCD[m,n]]