The solutions that your teacher found are the $4$ solutions to $z^4=16$. Here we will find these roots algebraically.
As you point out, the solutions should be of the form $z=2\cdot (-1)^{1/4}$, where "$(-1)^{1/4}$" stands for any fourth root of $-1$. Thus, we need to find the solutions to $z^4=-1,$ or equivalently, the solutions to $z^4+1=0$. Over the complex numbers, $-1$ is a square, namely $i^2=-1$, so $z^4+1 = z^4-(-1) = z^4-i^2$ and therefore $z^4+1$ factors as $(z^2-i)(z^2+i)$. In order to factor this expression completely, you need to find the square roots of $i$ and the square roots of $-i$.
Let us find the square roots of $i$, i.e., we are looking for a complex number $a+bi$ such that $(a+bi)^2=i$. Since $(a+bi)^2=a^2-b^2+2abi$, we conclude that $a^2-b^2=0$ (so $a=\pm b$) and $1=2ab=\pm 2b^2$. Thus, $a=b=\pm \frac{\sqrt{2}}{2}$. In other words, the square roots of $i$ are $z_1=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i,\quad \text{and} \quad z_2=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i.$ You can similarly find that the square roots of $-i$ are given by $z_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i,\quad \text{and} \quad z_4=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i.$ Therefore, it follows from our previous discussion that the roots of $z^4+16=0$ are $2z_1$, $2z_2$, $2z_3$ and $2z_4$, which are the solutions your teacher found.