It seems that those integers which suits \[n\equiv -1(mod p)\]will help.

No because p can be changed arbitraryly, it's not fixed.

Although I am the person who asked this question, I 've come up with this solution. Hope that anyone can verify it for me. Many thanks. Firstly, we will find out when the fraction is an integer. We have $$\frac{p^a-p^b}{p^c-p^d}=\frac{p^b(p^{a-b}-1)}{p^d(p^{c-d}-1)}$$So we can infer that $b\geq d$ and from the lemma undermentioned, we conclude that $c-d\mid a-b$. Lemma. For $a\geq 2 \in \mathbb{Z}$ and $m,n$ are positive integers, we have $$\gcd(a^m-1,a^n-1)=a^{\gcd(m,n)}-1$$. Now we're good to go, there exists nonnegative integers $x,y,z$ such that $$\frac{p^a-p^b}{p^c-p^d}=p^x\cdot \frac{p^{yz}-1}{p^y-1}$$. For the sake of clarity, for a prime $p$,denote a number $n$ to be $p$-good if there doesn't exist nonnegative integers $x,y,z$ such that $$n=p^x\cdot \frac{p^{yz}-1}{p^y-1}$$. Now it's easy to see that if this equation has integer root $(x,y,z)$ then either $p\mid n$ or $p\mid n-1$. It follows that $n$ is a $p$-good number for all prime $p> n$. (1) We will prove that for sufficiently large $k$, the number of the form $$2^k\cdot 11$$will be $p$-good for all $p< n$, then from (1), we obtain the satisfied number. Indeed, consider the funtion $$f(x,b)=\frac{x^b-1}{x-1},x,b\in\mathbb{Z}$$, then this funtion is strictly increasing with respect to $x$ and $b$. So, for all $x\geq 3$, we have $f(x,b)=\frac{x^b-1}{x-1}\geq \frac{3^b-1}{3-1}$. And now there exists $n_0$ sufficiently large such that, for all $b\geq n_0$ we have $f(x,b)=\frac{x^b-1}{x-1}\geq \frac{3^b-1}{3-1}>11b$. Pick two prime of the form $11k+1$, let's choose $23$ and $67$ then it's easy to see that, if we call $t=lcm(\varphi(23),\varphi(67))=66$ then $\gcd(t,10)=2$. Choose $k$ sufficiently large enough such that $2^k>n_0$, $k\equiv 0 \mod t$ and $k\equiv 2 \mod 10$ , since $\gcd(t,10)\mid 2-0$ thus this congruent equation has a root $k$. By doing this, we have $23\cdot 67\mid 2^k\cdot 11-1$, thus we have ensured that $n-1$ is not a prime number. We will prove that $n=2^k\cdot 11$ is the desired number. We observe that $11=\overline{1011}_2$, so $n$ is a $2$-good number. Consider the equation $$n=11^x\cdot \frac{11^{yz}-1}{11^y-1}$$Suppose this equation has integer root $(x,y,z)$, since $v_{11}(n)=1$, it follows that $x=1$ and $2^k= \frac{11^{yz}-1}{11^y-1}\equiv 1\mod 11$ but since $k\equiv 2\mod 10\Rightarrow 2^k\equiv 2^2\equiv 4\mod 11$, a contradiction. So $2^k\cdot 11$ is a $11$-good number. So far we've prove that $n$ is $p$ good for all $p\mid n$, thus we remained to consider $p\mid n-1$. Since $x=0$, we have to consider the equation: $$2^k\cdot 11= \frac{p^{yz}-1}{p^y-1}$$. Suppose this equation has root for $p\mid n-1$, of course $p\neq 2$, then $k=v_2(2^k\cdot 11)=v_2\left(\frac{p^{yz}-1}{p^y-1}\right)=v_2(z)$. Thus $z\geq 2^k>n_0$ and $2^k\cdot 11=2^{v_2(z)}\cdot 11\leq 11z$. Since $p^y\geq 3$, we infer that $$ \frac{p^{yz}-1}{p^y-1}\geq \frac{3^{z}-1}{3-1}>11z\geq 2^k\cdot 11$$, because $z>n_0$ which is a contradiction. So $n$ is a $p$-good number for all $p\mid n-1$. So $n$ is the desired number and it's easy to see that there are infinitely many ways to choose $k$. Hence proved.