3. Note that,
$$\begin{array}{l} 4 x^{4}+y^{4}-z^{2}+4 x y z \\ =\left(4 x^{4}+y^{4}+4 x^{2} y^{2}\right)-\left(4 x^{2} y^{2}+z^{2}-4 x y z\right) \\ =\left(2 x^{2}+y^{2}\right)^{2}-(2 x y-z)^{2} \\ =\left(2 x^{2}+y^{2}-2 x y+z\right)\left(2 x^{2}+y^{2}+2 x y-z\right) . \end{array}$$

Let $A=2 x^{2}+y^{2}-2 x y+z$,
$$B=2 x^{2}+y^{2}+2 x y-z \text {. }$$

Then $A+B=4 x^{2}+2 y^{2}$.
If we choose the values of $x$ and $y$ such that
$$A=4 x^{2}=4 \times 5^{2 n+2}, B=2 y^{2}=2 \times 2^{2 n} \text {, }$$

then $A B=2 \times 10^{2 n+2}$, and the sum of the digits of $A B$ must be 2.
Thus, for any integer $n \geqslant 1, x=5^{n+1}, y=2^{2 n}$, we have
$$\begin{aligned} & A=4 x^{2} \Leftrightarrow 2 x^{2}+y^{2}-2 x y+z=4 x^{2} \\ \Rightarrow & z=2 x^{2}+2 x y-y^{2} \\ = & 2 \times 5^{2 n+2}+10^{n+1}-4^{n} . \end{aligned}$$

Since $z$ is a positive integer,
$$4 x^{4}+y^{4}-z^{2}+4 x y z=2 \times 10^{2 n+2} \text {. }$$

Therefore, the conclusion holds.