It is clear that $(0, 0, 0, 0)$ is a solution of the equation. We will show that there exists no nonzero solution of the equation. Let us suppose the contrary, i.e. let $(x_0, y_0, z_0, t_0)$ be a solution of the equation with at least one nonzero coordinate $x_0^4 + 2y_0^4 + 4z_0^4 + 8t_0^4 = 16x_0 y_0 z_0 t_0$. It is clear that $x_0$ is divisible by $2$. Therefore we write it in the form $x_0 = 2x_1$ for some $x_1 \in \mathbb{Z}$, we substitute in the first equation and we divide by $2$, after which we get the equation $y_0^4 + 2z_0^4 + 4t_0^4 + 8x_1^4 = 16x_1 y_0 z_0 t_0$. Now the number $y_0$ is divisible by $2$ so $y_0 = 2y_1$ for some $y_1 \in \mathbb{Z}$. We again substitute in the last equation and we divide by $2$, after which we get $z_0^4 + 2t_0^4 + 4x_1^4 + 8y_1^4 = 16x_1 y_1 z_0 t_0$. Analogously, the number $z_0$ is divisible by $2$, so $z_0 = 2z_1$ for $z_1 \in \mathbb{Z}$. We substitute in the last equation and divide by $2$, after which we get the equation $t_0^4 + 2x_1^4 + 4y_1^4 + 8z_1^4 = 16x_1 y_1 z_1 t_0$. The number $t_0$ is divisible by $2$ i.e. $t_0 = 2t_1$ for $t_1 \in \mathbb{Z}$. After substitution and division by $2$ we get the equation $x_1^4 + 2y_1^4 + 4z_1^4 + 8t_1^4 = 16x_1 y_1 z_1 t_1$. We get that the quadruple $(x_1, y_1, z_1, t_1)$ is also a solution of the first equation. If we continue this procedure, we get quadruples of integer solutions of the first equation $(x_2, y_2, z_2, t_2)$, $(x_3, y_3, z_3, t_3)$, with the solution at the $n$-th step being $(x_n, y_n, z_n, t_n) = \left(\frac{x_0}{2^n}, \frac{y_0}{2^n}, \frac{z_0}{2^n}, \frac{t_0}{2^n}\right)$, which contradicts the fact the solutions are integer quadruples. Therefore the only solution is $(0, 0, 0, 0)$.