jmc

We can write the given equation as $$z^2-4z=-19+8i.$$Then $z^2-4z+4=-15+8i,$ so $(z-2{)}^2=-15+8i.$

Let $-15+8i=(a+bi{)}^2,$ where $a$ and $b$ are real numbers. Expanding, we get $$-15+8i=a^2+2abi-b^2.$$Setting the real and imaginary parts equal, we get $a^2-b^2=-15$ and $ab=4.$ Hence, $b=\frac{4}{a},$ so $$a^2-\frac{16}{a^2}=-15.$$Then $a^4-16=-15a^2,$ so $a^4+15a^2-16=0.$ This factors as $(a^2-1)(a^2+16)=0.$ Since $a$ is real, $a=\pm 1,$ which leads to $b=\pm 4.$ Thus, $$z-2=\pm (1+4i),$$Then $z=3+4i$ or $z=1-4i.$ Only $\boxed{3+4i}$ has an integer magnitude.