Hint If $(a, b, c, d)$ is a solution, substitution shows that so are the infinitely many tuples
$$(\lambda^{4 \cdot 5 \cdot 7} a, \lambda^{3 \cdot 5 \cdot 7} b, \lambda^{3 \cdot 4 \cdot 7} c, \lambda^{3 \cdot 4 \cdot 5} d), \qquad \lambda \in \Bbb Z .$$
So, it suffices to find a single nonzero solution,
It's easy to find a nonzero solution (e.g., $(1,0,0,1)$), and it's not too much harder to find a solution in positive integers: Using the decomposition $2^{m + 2} = 2^{m + 1} + 2^{m + 1} = 2^{m + 1} + 2^m + 2^m$ (this observation is used in achille hui's solution, too), substituting shows that $$(2^{\alpha}, 2^{\beta}, 2^{\gamma}, 2^{\delta})$$ is a solution if, for example, $$3 \alpha = 5 \gamma = m, \qquad 4 \beta = m + 1, \qquad 7 \delta = m + 2$$
for some integer $m$. The first condition implies $15 \mid m$, and then the second condition gives $m = 15(4 k + 1)$ for some integer $k$. The smallest positive integer solution that also satisfies the last condition is then $k = 1$, or $m = 75$. Substituting gives that the corresponding solution is
$$(2^{25}, 2^{19}, 2^{15}, 2^{11}) = (33\,554\,432, 524\,288, 32\,768, 2\,048) .$$