So far we've had only one possible value for p (within computer range), so I've been trying to prove that no others exist. Will this do?

In factorised form the expression looks like this:
   (2^1994)(1 + 2^4 + 2^5 + 2^6 + 2^8 + 2^(p - 1994))
= (2^997)²(369 + 2^q)    where q = p - 1994

This is a perfect square iff 369 + 2^q is a perfect square
i.e.   369 + 2^q = u²       for integers q and u       (1)

It follows that  2^q = u²  (mod 3)

Now, 2^q can only be 1 or 2 (mod 3) and u² can only be 0 or 1 (mod 3) so, for a solution, 2^q = u² = 1 (mod 3), but this can only happen when q is even.

So, let q = 2r  then (1) becomes   369 + 2^(2r) = u², and can be factorised to give
              3 x 3 x 41 = (u - 2^r)(u + 2^r)

This allows only three possibilities:
(a) u - 2^r = 1 and u + 2^r = 369, giving 2^r = 184, so no integer root
(b) u - 2^r = 3 and u + 2^r = 123, giving 2^r = 60, so no integer root
(c) u - 2^r = 9 and u + 2^r = 41, giving 2^r = 16.
This last possibility gives the only solution: r = 4, u = 25.

So p - 1994 = q = 2r = 8, giving  p = 2002 is the only solution.

But there must be a neater way?..