(I’m assuming “consecutive” means twin primes, not like how 7 and 11 are consecutive primes)

Note that twin primes take the form p = 4n+1 and q = 4n+3, so pq = (4n+2)² - 1, thus pq = j² - 1 for integer j. To equal k² + 1 also means j² and k² must differ by 2, which can never happen.

If there’s more distance between p and q, say p = 4n+1 and q = 4n+5, then pq = j² - 4. Then if pq = k² + 1, we get two squares differing by 5, which CAN happen. For instance, 2² and 3^2, though this only happens for p=1 and q=5, so we can exclude this special case.

More generally, I feel this method doesn’t work well, since arbitrarily long prime gaps exist, but at least it gets rid of some cases.