Solution:
Given $\sin^2 A + \sin^2 B = \sin (A + B)$.

We know that $\sin^2 A = 1 - \cos^2 A$ and $\sin^2 B = 1 - \cos^2 B$, but let's try to use the sum-to-product identities.

Recall that:
$$\sin^2 A + \sin^2 B = \frac{1 - \cos 2A}{2} + \frac{1 - \cos 2B}{2} = 1 - \frac{\cos 2A + \cos 2B}{2}$$
So the equation becomes:
$$1 - \frac{\cos 2A + \cos 2B}{2} = \sin (A + B)$$
Bring $\sin (A + B)$ to the left:
$$1 - \sin (A + B) = \frac{\cos 2A + \cos 2B}{2}$$
Multiply both sides by $2$:
$$2 - 2\sin (A + B) = \cos 2A + \cos 2B$$
But $\cos 2A + \cos 2B = 2 \cos \frac{2A + 2B}{2} \cos \frac{2A - 2B}{2} = 2 \cos (A + B) \cos (A - B)$.

So:
$$2 - 2\sin (A + B) = 2 \cos (A + B) \cos (A - B)$$
Divide both sides by $2$:
$$1 - \sin (A + B) = \cos (A + B) \cos (A - B)$$
Bring all terms to one side:
$$1 - \sin (A + B) - \cos (A + B) \cos (A - B) = 0$$
Let $S = A + B$ and $D = A - B$.

So:
$$1 - \sin S - \cos S \cos D = 0$$
But $\cos S \cos D = \frac{1}{2} [\cos(S - D) + \cos(S + D)] = \frac{1}{2} [\cos(A + B - (A - B)) + \cos(A + B + (A - B))] = \frac{1}{2} [\cos(2B) + \cos(2A)]$

But this brings us back to the earlier form. Let's try another approach.

Let us suppose $A + B = x$.

Then $\sin^2 A + \sin^2 B = \sin x$.

But $\sin^2 A + \sin^2 B = 1 - \frac{\cos 2A + \cos 2B}{2}$ as above.

So:
$$1 - \frac{\cos 2A + \cos 2B}{2} = \sin x$$
So:
$$\cos 2A + \cos 2B = 2(1 - \sin x)$$
But $2A + 2B = 2x$, so $\cos 2A + \cos 2B = 2 \cos x \cos(A - B)$.

Therefore:
$$2 \cos x \cos(A - B) = 2(1 - \sin x)$$
Divide both sides by $2$:
$$\cos x \cos(A - B) = 1 - \sin x$$
But $A$ and $B$ are acute, so $0 < A < \frac{\pi}{2}$, $0 < B < \frac{\pi}{2}$, so $0 < x < \pi$.

Suppose $x = \frac{\pi}{2}$.
Then $\sin x = 1$, $\cos x = 0$.
So left side: $0 \cdot \cos(A - B) = 0$, right side: $1 - 1 = 0$.
So equality holds.

Suppose $x < \frac{\pi}{2}$.
Then $\sin x < 1$, $\cos x > 0$.
So $1 - \sin x > 0$, $\cos x > 0$, so $\cos(A - B) = \frac{1 - \sin x}{\cos x}$.
But $|A - B| < x < \frac{\pi}{2}$, so $\cos(A - B) > 0$.
But $\cos(A - B) \leq 1$, so $\frac{1 - \sin x}{\cos x} \leq 1$.
So $1 - \sin x \leq \cos x$.
But $1 - \sin x - \cos x \leq 0$.
But for $0 < x < \frac{\pi}{2}$, $1 - \sin x - \cos x > 0$ for small $x$, so possible only at $x = \frac{\pi}{2}$.

Similarly, for $x > \frac{\pi}{2}$, $\cos x < 0$, $1 - \sin x < 0$, but $\cos(A - B) > 0$, so left side negative, right side negative, but $\cos(A - B) = \frac{1 - \sin x}{\cos x}$, but $\cos(A - B) > 0$, so $\frac{1 - \sin x}{\cos x} > 0$, but both negative, so possible only if $x = \frac{\pi}{2}$.

Therefore, the only solution is $A + B = \frac{\pi}{2}$.