14.1.2 ** Find all pairs of positive integers $(a, b)$, such that: $a^{2}+b+1$ is a power of a prime, $a^{2}+b+1$ divides $b^{2}-a^{3}-1$, but does not divide $(a+b-1)^{2}$.
Parse $b^{2}-a^{3}-1=-a\left(a^{2}+b+1\right)+(b+1)(a+b-1)$, so $a^{2}+b+1$ divides $(b+1)(a+b-1)$.

Let $a^{2}+b+1=p^{k}, p$ be a prime, then $p^{k} \mid(b+1)(a+b-1)$. Since $p^{k} \nmid(a+b-1)^{2}$, it must be that $p \mid(b+1)$. Also, $a^{2}+b+1>b+1$, so $p^{k} \nmid(b+1), p \mid(a+b-1)$. Thus, $p$ divides $a^{2}+b+1-(b+1)=a^{2}, p$ divides $a . p$ divides $a+b+1-(a+b-1)=2 . p=2$.

Let $a=2^{s} a_{1}, b+1=2^{t} b_{1}, s 、 t$ be positive integers, $a_{1} 、 b_{1}$ be positive odd numbers. Since $2^{2 s} a_{1}^{2}+2^{t} b_{1}-2^{k}$, it follows that $k>2 s, k>t .2^{2 s}$ divides $2^{k}$, thus $2^{2 s}$ divides $2^{t}$, i.e., $2 s \leqslant t$. Similarly, $t \leqslant 2 s$. Therefore, $t=2s. a_{1}^{2}+b_{1}=2^{k-2 s} \geqslant 2$. $a+b-1=2^{s} a_{1}+2^{2 s} b_{1}-2$.

If $s>1$, then the power of 2 in $a+b-1$ is 1. From $a^{2}+b+1$ divides $(b+1)(a+b-1)$, we get $k \leqslant 2 s+1$. Thus, $k=2 s+1$ and $a_{1}^{2}+b_{1}=2, a_{1}=b_{1}=1$. In this case, $a=2^{s}, b=2^{2 s}-1$. If $s=1$, then $a+b-1=2\left(a_{1}-1\right)+2^{2} b_{1}$. Since $a_{1}$ is odd, $a_{1}-1$ is even, and the power of 2 in $a+b-1$ is at least 2. But $a^{2}+b+1$ divides $(b+1)(a+b-1)$ but does not divide $(a+b-1)^{2}$, indicating that the power of 2 in $a+b-1$ is less than the power of 2 in $b+1$, i.e., less than $2 s=2$, a contradiction!
Therefore, the solution to this problem is $a=2^{s}, b=2^{2 s}-1, s=2,3,4, \cdots$.