Note that $11(a-b)|a^3b-b^4$, or $11|b(a^2+ab+b^2)$. This implies $11|b$, since if $11|a^2+ab+b^2$ then $11|a,11|b$. Let's get some trivial cases out of the way: when $b=0$, $a=0$. When $a=0$, $b=0,11$. Otherwise, $a,b\neq 0$. Then they must be of the same sign (since $a^3b>0$). The key idea for bounding is that $a^3b$ is always bounded by $a^2b^2$. If we imagine that we started from $a=b$ and then slowly shifted $a$ away, then $a^3b$ would be the most "extreme", followed by $a^2b^2$, followed by $(b^2+11(a-b))^2$, so something should go wrong with the bound. Firstly, if $|a|>|b|$, then $(b^2+11(a-b))^2=a^3b>a^2b^2$. Thus $|b|^2+11|a-b|\ge |b^2+11(a-b)|>|ab|$ and so $11|a-b|>|b|\cdot |a-b|$, contradiction. If $|a|<|b|$, then $(b^2+11(a-b))^2=a^3b<a^2b^2$. But $b^2>11(b-a)$, so $b^2+11(a-b)<|ab|$, and $|b|\cdot |b-a|<11(b-a)$. Again, contradiction. Hence, $a=b$. So either $a=b$ or $(a,b)=(0,11)$.

Unfortunately expanding was all I could see... We first see that $a=b$ is a solution. Now assume $a\neq b$. Let $a-b=x$ Subbing into original equation, we get $22b^2+121x=3b^3+3b^2+bx^2$. Note in the process we divided by $x$, which is possible since $a\neq b$. Subbing back $x=a-b$ and moving all terms to the LHS, we get $$a^2b+a(b^2-121)+(b^3-22b^2+121b)=0.$$ Hence $\Delta = -3b^4+8\cdot 11b^3-6\cdot 11^2b^2+11^4$, which (surprisingly?) factorizes nicely into $$(b-11)^2(b-11)(-3b-11).$$ Since $\Delta$ is positive, we see that $(b-11)(-3b-11)$ is positive, from which we get $-3\leq b \leq 11$. It is now easy to check that the only solution is when $b=11$ and $(a,b)=(0,11)$ is a solution, as well as $a=b$ from above.

The answers are $a=b$ and $(a,b)=(0,11)$. Let's say that we have that $a,b \neq 0$ Then we set $a=dx$ and $b=dy$, wherer $x$ and $y$ are relatively coprime. Then we have that: $$(dy^2+11(x-y))^2=d^2x^3y$$this implies that $y \mid dy^2+11(x-y)$ and this implies that $y \mid 11$. Thus we conclude that $y \in \{-11,-1,1,11\}$, and after a lengthy bash we get that $x=y=1$. Now let's see when one of then is equal to $0$. Easily when $b=0$, we get that $a=0$. If $a=0$, we get that $b^2-11b=0$, this implies that $b=11$ and $b=0$.