Rearrange the equation as $x^2 - y^2 = -16$, which further simplifies to $(x-y)(x+y) = -16$.
Now consider the factors of $-16$. Since $x-y$ and $x+y$ are integers, both $x$ and $y$ must be even or both must be odd.
Solution: The given equation is $x^2 + 16 = y^2$, which can be rearranged as
\[
x^2 - y^2 = -16 \implies (x-y)(x+y) = -16.
\]
As we know that both $x$ and $y$ are integers and hence $x-y$ and $x+y$ are also integers. Let $-16 = a \times b$ where $a$ and $b$ are integers, then the possible values of $a$ and $b$ are
\begin{align*}
(\pm 1, \mp 16), (\pm 16, \mp 1), (\pm 2, \mp 8), (\pm 8, \mp 2), (\pm 4, \mp 4).
\end{align*}
If $x - y = a$ and $x + y = b$, then
\[
x = \frac{a+b}{2} \quad \text{ and } \quad y = \frac{b-a}{2}.
\]
As $x, y \in \mathbb{Z} $, so $a$ and $b$ must have the same parity. Thus, we have the following pairs of $(a,b)$:
\[
(\pm 2, \mp 8), (\pm 8, \mp 2), (\pm 4, \mp 4).
\]
Hence, the number of integral solutions to the equation $x^2 + 16 = y^2$ is $6$.