EDIT
There seems to be a lot of non-harmonic solutions.
Here are couple of one parameter family of non-harmonic solutions.
$b = a(a-1)$, $d = a^2 - a + 1$ and $c = (a-1)(2a^2 - a +1)$ which relies on the identity $a^4 + \left( a(a-1)\right)^4 + \left((a-1)(2a^2-a+1) \right)^2 = (a^2 - a + 1)^2$ You could scale these up appropriately i.e. $\left(ka,ka(a-1),k^2(a-1)(2a^2 - a + 1),k \left(a^2-a+1 \right) \right),$ to get other solutions.
$b = a(a+1), d = a^2 + a + 1$ and $c = (a+1)(2a^2+a+1)$ which relies on the identity $a^4 + \left( a(a+1)\right)^4 + \left((a+1)(2a^2+a+1) \right)^2 = (a^2 + a + 1)^2$ You could again scale these up appropriately i.e. $\left(ka,ka(a+1),k^2(a+1)(2a^2 + a + 1),k \left(a^2+a+1 \right) \right),$ to get other solutions.
You could also take your harmonic solution $(a,a+1,a(a+1)(a^2 + a + 2),a^2+a+1)$ and scale appropriately, i.e. $\left(ka,k(a+1),k^2a(a+1)(a^2 + a + 2),k \left(a^2+a+1 \right) \right),$ to get other solutions.
Yes. Below is a one parameter family of infinite solutions.
$b = a+1$, $d = a^2+a+1$, $c = (a^2+a+1)^2-1$.
$b^4 = (a+1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1$
$d^4 = (a^2 + a +1)^4 = 1+4 a+10 a^2+16 a^3+19 a^4+16 a^5+10 a^6+4 a^7+a^8$
Hence, \begin{align} d^4 - b^4 - a^4 & = 4 a^2+12 a^3+17 a^4+16 a^5+10 a^6+4 a^7+a^8\\ & = a^2 \left(4 +12 a+17 a^2+16 a^3+10 a^4+4 a^5+a^6 \right)\\ & = a^2 (a+1)^2 (a^2 + a + 2)^2 \end{align} Hence, choose $c = a(a+1)(a^2+a+2) = (a^2 + a + 1 -1)(a^2 + a + 1 +1) = \left( \left(a^2 + a + 1 \right)^2 -1 \right)$