For any natural number $u$, there is a natural number $t$ such that $ut + 1$ and $ut + t + 1$ are both prime. So we get a solution of the equation
$$au - b(u+1) = -1$$
with prime numbers $a$ and $b$ by setting $a := ut + t + 1 , b := ut + 1$ :
$$(ut+t+1)u - (ut+1)(u+1) = u^2t+ut+u-u^2t-ut-u-1 = -1$$
If the conjecture is true, then for any natural number $n$, there is a pair $(a,a+1)$ of consecutive squarefree numbers with exactly $n$ distinct prime factors.
I checked the conjecture with PARI and for $u\le10^6$, it is true. The largest number t necessary to produce the prime pair is $3420$ for the number $829123$ upto $u = 10^6$