Let $b = 4d$, then
\[ 4ad - a - d = c^2 \]It is pretty well known that $4ab - a - b$ can't be a square. So there are no such triples.
Let $4ab - a - b = c^2$. Then
\[ (4a - 1)(4b - 1) = 4c^2 + 1 \]Note that $LHS$ has a prime factor of $3$ mod $4$. While $RHS$ can't have any factor of 3 mod 4.