Nice problem!

Let be distinct primes of the form , and let ; using the Chinese Remainder Theorem, we can solve the congruences

simultaneously. Now suppose for where , and for where . Let where

(set if ). I claim that for all . First, suppose for some ; since , we must have which is clearly false. On the other hand, if for some , since we have we must have which is also false.

Thus the consecutive integers are each divisible by a prime of the form only once, and hence none of them is a sum of two squares. By taking to be as large as we wish, we solve the problem.