14. Find all positive integers $n$ and prime numbers $p$ that satisfy $17^{n} \cdot 2^{n^{2}}-p=\left(2^{n^{2}+3}+2^{n^{2}}-1\right) \cdot n^{2}$.
When $p=2$, $17^{n} \cdot 2^{n^{2}}-2 \equiv 2(\bmod 4)$, while $\left(2^{n^{2}+3}+2^{n^{2}}-1\right) \cdot n^{2}$ is either odd or a multiple of 4, leading to a contradiction.
If $p$ is an odd prime, then $n$ is odd. When $n=1$, $34-p=17 \Rightarrow p=17$, which satisfies the condition;
When $n \geqslant 3$, $n^{2} \equiv 1(\bmod 8), 2^{4} \equiv-1 \Rightarrow 2^{8} \equiv 1(\bmod 17)$, thus $2^{n^{2}+3} \equiv-1(\bmod 17), 2^{n^{2}} \equiv 2$ $(\bmod 17) \Rightarrow 17 \mid\left(2^{n^{2}+3}+2^{n^{2}}-1\right) \Rightarrow p=17$.
The original equation can be transformed into $17\left(17^{n-1} \cdot 2^{n^{2}}-1\right)=9 n^{2} \cdot 2^{n^{2}}-n^{2}$. Clearly, when $n \geqslant 3$, $17^{n-1}>9 n^{2}$, then $17\left(17^{n-1} \cdot 2^{n^{2}}-1\right) > 17\left(9 n^{2} \cdot 2^{n^{2}}-1\right)$, and $9 n^{2} \cdot 2^{n^{2}}-n^{2}<9 n^{2} \cdot 2^{n^{2}}-1$, leading to a contradiction.
In conclusion, the original equation has a unique solution $(p, n)=(17,1)$.