We show that there are no such matrices. Notice that $A^{4}+4 A^{2} B^{2}+16 B^{4}$ can factorized as $$A^{4}+4 A^{2} B^{2}+16 B^{4}=\left(A^{2}+2 A B+4 B^{2}\right)\left(A^{2}-2 A B+4 B^{2}\right)$$ Let $C=A^{2}+2 A B+4 B^{2}$ and $D=A^{2}-2 A B+4 B^{2}$ be the two factors above. Then $$\operatorname{det} C \cdot \operatorname{det} D=\operatorname{det}(C D)=\operatorname{det}\left(A^{4}+4 A^{2} B^{2}+16 B^{4}\right)=\operatorname{det}(2019 I)=2019^{n}$$ The matrices $C, D$ have integer entries, so their determinants are integers. Moreover, from $C \equiv D(\bmod 4)$ we can see that $$\operatorname{det} C \equiv \operatorname{det} D \quad(\bmod 4)$$ This implies that $\operatorname{det} C \cdot \operatorname{det} D \equiv(\operatorname{det} C)^{2}(\bmod 4)$, but this is a contradiction because $2019^{n} \equiv 3$ $(\bmod 4)$ is a quadratic nonresidue modulo 4. Solution 2. Notice that $$A^{4} \equiv A^{4}+4 A^{2} B^{2}+16 B^{4}=2019 I \quad \bmod 4$$ so $$(\operatorname{det} A)^{4}=\operatorname{det} A^{4} \equiv \operatorname{det}(2109 I)=2019^{n} \quad(\bmod 4)$$ But $2019^{n} \equiv 3$ is a quadratic nonresidue modulo 4, contradiction.