We shall prove that $a = b = c = d = 0$. It is easy to see that if one of the numbers $a, b, c, d$ equals $0$, then the others equal $0$, too. Indeed, if $b = 0$, $a \neq 0$, then $\sqrt{3} = \pm\frac{c}{a^2}$ is a rational number, a contradiction. The case $a = 0$, $b \neq 0$ is impossible by the same reasoning. Let $c = 0$. Then $n = -b > 0$ and $3a^4 = 2n^3$, $3a^6 \ge n^5$. Hence $2a^2 \ge n^2$ and therefore $2n^3 = 3a^4 \ge \frac{3n^4}{4}$. Then $n \le \frac{8}{3}$, i.e. $n = 1$ or $n = 2$, a contradiction. Analogously $d \neq 0$.

Let now $a, b, c, d \neq 0$. Adding both equations and using that $3$ divides $b^5 + 2b^3 = b^3(b-1)(b+1) + 3b^3$, we get that $3|c^2+d^2$, i.e. $3|c, d$. Hence $3|a, b$. Let $a = 3^\alpha a_1$, $b = 3^\beta b_1$, $c = 3^\gamma c_1$, $d = 3^\delta d_1$, where $\alpha, \beta, \gamma, \delta \ge 1$ and $3 \nmid a_1, b_1, c_1, d_1$. Then the system can be written in the form
$$(1) \qquad \left| \begin{array}{l} 3^{4\alpha+1}a_1^4 + 3^{3\beta}2b_1^3 = 3^{2\gamma}c_1^2 \\ 3^{6\alpha+1}a_1^6 + 3^{5\beta}b_1^5 = 3^{2\delta}d_1^2 \end{array} \right.$$
We shall use the following trivial fact. If $3^k p + 3^l q = 3^m r$ and $3 \nmid p, q, r$, then at least two of the numbers $k, l, m$ are equal. This and (1) imply that $4\alpha + 1 = 3\beta$ or $3\beta = 2\gamma$, and $6\alpha + 1 = 5\beta$ or $5\beta = 2\delta$. Then it is easy to see that $3\beta = 2\gamma$ and $5\beta = 2\delta$. Now (1) is equivalent to
$$\left| \begin{array}{l} 3^{4\alpha+1-2\gamma}a_1^4 + 2b_1^3 = c_1^2 \\ 3^{6\alpha+1-2\delta}a_1^6 + b_1^5 = d_1^2 \end{array} \right.$$
Since $4\alpha + 1 - 2\gamma > 0$ and $6\alpha + 1 - 2\delta > 0$, adding the last two equations, we conclude as above that $3|c_1, d_1$, a contradiction.