Solution:

The equation is symmetric in $a$ and $b$, so we can restrict ourselves to considering solutions with $a \geq b$. There are no solutions with $a$ and $b$ both positive: indeed in that case we would have $a^{3}+b^{3}+3 a b \geq 1+1+3=5>1$, a contradiction.

If at least one of $a$ and $b$ is equal to zero, say $b$, then substituting into the equation we find $a^{3}=1$, that is $a=1$, giving the two solutions $(1,0)$ and $(0,1)$.

If then $a$ and $b$ are both strictly less than $0$, then (still assuming $b \leq a<0$) we find $a^{3}+b^{3}+3 a b < b\left(b^{2}+3 a\right)$. If we had $b \leq -3$, then we would also have $b^{2}+3 a \geq -3 b+3 a=3(a-b) \geq 0$ and therefore $b\left(b^{2}+3 a\right)$ could not be positive. It would follow that $a^{3}+b^{3}+3 a b < b\left(b^{2}+3 a\right) \leq 0$, and hence $(a, b)$ cannot be a solution if $b \leq -3$. We must therefore check the cases $b=-2$ and $b=-1$, which (recalling that we are assuming $a<0$) lead to the unique solution $(a, b)=(-1,-1)$.

It remains for us to determine the solutions with $a>0$ and $b<0$: let us set $c=-b$, so that $c$ is a positive integer. The equation can be rewritten in the two equivalent forms
$$3 a c+1=a^{3}-c^{3}, \quad a^{3}=c^{3}+3 a c+1 ;$$
from the second expression we deduce that $a>c$, while from the first we find $3 a c+1=(a-c)\left(c^{2}+c a+a^{2}\right)$. Since $a-c \geq 1$ we then have
$$3 a c+1=(a-c)\left(c^{2}+c a+a^{2}\right) \geq c^{2}+c a+a^{2},$$
from which $1 \geq c^{2}-2 c a+a^{2}$, that is $(a-c)^{2} \leq 1$. Since we have already observed that $a-c$ is positive, it must be that $a-c=1$, and substituting $a=c+1$ we find
$$c^{3}+3 c^{2}+3 c+1=c^{3}+3 c(c+1)+1,$$
an equation that is satisfied for any value of $c$. Recalling that at the beginning we had assumed $a \geq b$ we then see that the solutions of the proposed equation are the pair $(a, b)=(-1,-1)$ and the infinitely many pairs $(a, b)=(a,-c)=(c+1,-c)$ for every integer value of $c$: those with $c \geq 0$ are those for which $a \geq b$, and those with $c<0$ are the symmetric ones with $a<b$.

Second Solution: We observe that
$$a^{3}+b^{3}-1+3 a b=a^{3}+b^{3}+(-1)^{3}-3 a b \cdot(-1)=(a+b-1)\left(a^{2}+b^{2}+1-a b+a+b\right),$$
so the pairs of the form $(n, 1-n)$ are solutions for every integer $n$.

Multiplying the second factor by $4$, moreover, we can write
$$4 a^{2}+4 b^{2}+4-4 a b+4 a+4 b=(2 a-b+1)^{2}+3(b+1)^{2},$$
which, being a sum of squares, vanishes only when $(2 a-b+1)=(b+1)=0$, that is when $a=b=-1$.