In fact, $a$ also cannot be odd. Recall that $x^2 \equiv 0, 1 \pmod{4}$ for all integers $x$, so if $a$ is odd, then $5a^4 + a^2 \equiv 2 \pmod{4}$ and $b^4 + 3b^2 + 4 \equiv 0 \pmod{4}$, a contradiction since $4 \mid b^4 + 3b^2 + 4$ but $4 \nmid 5a^4 + a^2$. Now it suffices to prove $a \neq 2$. Assume $a = 2$, then $b^4 + 3b^2 + 4 \mid 84$, and by checking all possible cases we see that $b$ has no solution, done.