5. Let $x$ and $y$ be positive integers such that
$$
\sqrt{x-116}+\sqrt{x+100}=y \text {. }
$$

Find the maximum value of $y$.
Prompt: By analogy with Example 6, we can prove that $\sqrt{x-116}$ and $\sqrt{x+100}$ are both natural numbers.
Let $\sqrt{x-116}=a, \sqrt{x+100}=b$. Then $b^{2}-a^{2}=216 \Rightarrow (b+a)(b-a)=216$.
Also, $b+a \equiv (b-a) \pmod{2}$, and since 216 is even, both $b-a$ and $b+a$ are even and positive.
Thus, $b-a \geqslant 2$.
Then $y=b+a \leqslant 108$.
When $\left\{\begin{array}{l}b+a=108, \\ b-a=2\end{array}\right.$, i.e., $\left\{\begin{array}{l}b=55, \\ a=53,\end{array}\right.$, i.e., $x=53^{2}+116$, $y$ reaches its maximum value.