Solution:
First Solution. Write the equation in the form
$$x^{2}(x+a)^{2}>4 x-3$$
Then for $x=1$ we get $(a+1)^{2}>1$, i.e. $a>0$ or $a<-2$. If $a<-2$, then $x=-a$ gives a contradiction $0>-4 a-3$. Thus, $a>0$.

Conversely, if $a>0$, then (1) is satisfied for all $x$. Indeed, when $x \leq 0$ this is obvious and when $x>0$ we have $x^{2}(x+a)^{2}>x^{4} \geq 4 x+3$, since the later inequality is equivalent to $(x-1)^{2}\left((x+1)^{2}+2\right) \geq 0$.

Second Solution. It follows from (1) that we have to find all $a$, for which
$$\text{a<- 4 x-3 x -x=f(x)}$$
for all $x \geq \frac{4}{3}$ or
$$a>\frac{\sqrt{4 x-3}}{x}-x=g(x)$$
for all $x \geq \frac{4}{3}$.

The first case is impossible since $\lim _{x \rightarrow+\infty} f(x)=-\infty$. The maximum of the function $g$ equals $0$ (and it is attained for $x=1$ ). Therefore the answer is $a>0$.