2.4. (SRP, 81). Solve the equation

$$
x^{6}+3 x^{3}+1=y^{4}
$$

in integers.
2.4. Let the equation be satisfied by some pair of numbers $x, y \in Z$. Suppose that $x>0$. Then we have

$$
\left(x^{3}+1\right)^{2}=x^{6}+2 x^{3}+1<x^{6}+3 x^{3}+1=y^{4}<x^{8}+4 x^{3}+4=\left(x^{3}+2\right)^{2}
$$

from which we obtain that the number $\boldsymbol{y}^{2}$ cannot be an integer, since

$$
x^{3}+1<y^{2}<x^{3}+2
$$

A similar contradiction is obtained when $x \leq-2$. Indeed, in this case $x^{3}+3<0$, so we have

$$
\left(x^{3}+2\right)^{2}=x^{6}+4 x^{3}+4<x^{6}+3 x^{3}+1=y^{4}<x^{6}+2 x^{3}+1=\left(x^{3}+1\right)^{2}
$$

from which we obtain the relations

$$
-\left(x^{3}+2\right)=\left|x^{3}+2\right|<y^{2}<\left|x^{3}+1\right|=-\left(x^{3}+1\right)
$$

which are not satisfied for any integer value of $y$. Further, when $x=-1$, the equation transforms into the equality $-1=y^{4}$, which is impossible. Finally, when $x=0$, we have the equality $1=y^{4}$. Thus, we obtain the answer: $x=0, y= \pm 1$.