$$

\left(16 x^{200}+1\right)\left(y^{200}+1\right)=16(x y)^{100}, x, y \in \mathbb{R}

$$

Let $x^{100}=a$ and $y^{100}=b$, where $a, b$ are non-negative reals. So, we have.

$$

\begin{aligned}

\left(16 a^{2}+1\right)\left(b^{2}+1\right) &=16 a b \\

16 a^{2} b^{2}+16 a^{2}+b^{2}+1 &=16 a b

\end{aligned}

$$

Applying AM-GM inequality on the terms $16 a^{2} b^{2}, 16 a^{2}, b^{2}, 1$ :

$$

\frac{16 a^{2} b^{2}+16 a^{2}+b^{2}+1}{4} \geq \sqrt[4]{16 a^{2} b^{2} \cdot 16 a^{2} \cdot b^{2} \cdot 1}=4 a b

$$

$$

\Longrightarrow 16 a^{2} b^{2}+16 a^{2}+b^{2}+1 \geq 16 a b

$$

Since $16 a^{2} b^{2}+16 a^{2}+b^{2}+1=16 a b$, from the equality condition of AM-GM inequality,

$$

\begin{array}{c}

16 a^{2} b^{2}=16 a^{2}=b^{2}=1 \Longrightarrow(a, b)=\left(\frac{1}{4}, 1\right) \\

\therefore(x, y) \in\left\{\left(\sqrt[100] \frac{1}{4}, 1\right),\left(\sqrt[100]{\frac{1}{4}},-1\right),\left(-\sqrt[100]{\frac{1}{4}}, 1\right),\left(-\sqrt[100]{\frac{1}{4}},-1\right)\right\}

\end{array}

$$

Therefore, there are 4 pairs of reals $(x, y)$ that satisfy the given equation.