You only need to check enough digits to find an odd one. An odd digit appears in the low order (< 46 up to a high level) digits for the first quadrillion cases so you only need to compute 2^n mod 10^d where d is big enough to be safe. I used d=60 in my computations to take this to 10^15 candidates (with no additional terms found).
Even that doesn't seem so bad though, it's on the order of 10^16 total digits to check in the worst case, and far fewer in practice.
Maybe someone here can run a program overnight and increase the bound by another few orders of magnitude, or disprove the hypothesis?