I have never studied Diophantine equations so I cannot say if there is a simple way to deal with this problem.

\(\displaystyle a, \ b, \ c, \ d \in \mathbb Z^+,\ a + b = cd \text { and } c + d = ab. \)

\(\displaystyle a + b = cd \implies b = cd - a.\)

\(\displaystyle \therefore c + d = ab = a(cd - a) = acd - a^2 \implies a^2 - acd + c + d = 0 \implies\)

\(\displaystyle a = \dfrac{cd \pm \sqrt{(cd)^2 - 4(1)(c + d)}}{2} = \dfrac{cd \pm \sqrt{c^2d^2 - 4c - 4d}}{2}.\)

That implies that \(\displaystyle c^2d^2 - 4c - 4d = e^2, \text { such that } \sqrt{e^2} \in \mathbb Z.\)

You found a solution at c = 2 and d = 3. So cd = 6, c^2d^2 - 4c - 4d = 36 - 8 - 12 = 16. And this in turn gives (6+4)/2 = 5 and (6-4)/2 = 1 as solutions. So yes, you can use a spread sheet to generate additional answers, but a looper program might be better. Denis is good at those looper programs. I did a quick and dirty, and it appears that you have found the only examples in the smallest 25 positive integers, but I did not test my program intensively.

EDIT: Because of symmetry, 1 and 5 are also solutions, but I did not count those as different from 2 and 3. Probably wrong of me. In that case, you get solutions that permute the integers involved.