Trivial. Observe that \[ b^4+3b^2+4=(b^2-b+2)(b^2+b+2)\implies 4\mid a^2(5a^2+1). \]Now if $a$ is odd, then $4\mid 5a^2+1$, absurd. So $a=2$ is the only possibility, which clearly does not work as it forces $b\le 2$, which is easily seen to be impossible.