Suppose that $a = p+1, \ b = q + 1, \ c = r+1$ where $p,q,r \geqslant 2$ are all prime numbers. We have $\left(p+1 \right) \left(q + 1 \right) \left(r+1 \right) = k^2 +1.$ If $p,q,r > 2,$ then $4 \mid k^2 + 1 \Rightarrow k^2 \equiv 3 \ (\text{mod 4}),$ contradiction. If $r = 2,$ we will have $k^2 \equiv 2 \ (\text{mod 3}),$ contradiction.

Suppose that at least one of $a,b,c$ is 3. Then $k^2\equiv 2 \pmod 3$, contradiction. Thus $a,b,c>3$. Assume that all $a-1,b-1,c-1$ are primes. Then $a,b,c$ are all even so $k^2\equiv7\pmod8$, contradiction.