Solution:
The only solution is $(a, b, c, p) = (3, 1, 2, 3)$. First, factor the right hand side. This gives us
$$2^{a}p^{b} = (p + 1)((p + 2)^{c - 1} + (p + 2)^{c - 2} + \dots +(p + 2) + 1).$$
Since $\gcd (p, p + 1) = 1$ it must be the case that $p + 1 = 2^{x}$ for some positive integer $x \leq a$ and so $p = 2^{x} - 1$ and $p + 2 = 2^{x} + 1$. Now for $x \geq 3$, $2^{x} + 1$ is not prime if $x$ is odd (since it is $0 \bmod 3$) and $2^{x} - 1$ is not prime if $x$ is even (since it is $0 \bmod 3$). This means $x \leq 2$, and the only admissible such $x$ is $x = 2$ since otherwise $p$ is not prime. So, the original equation becomes
$$2^{a}3^{b} = 5^{c} - 1.$$
Now $3|(5^{c} - 1)$, so evaluating $5^{c} - 1$ mod $3$ gives that $c = 2d$ for some positive integer $d$ and hence
$$2^{a}3^{b} = (5^{d} - 1)(5^{d} + 1).$$
Observe $5^{d} - 1$ and $5^{d} + 1$ are both even and have greatest common divisor $2$ because they are $2$ apart. Since $4|(5^{d} - 1)$ this implies $5^{d} - 1 = 2^{a - 1}$ and $5^{d} + 1 = 2\cdot 3^{b}$. Now, $3$ is not a factor of $5^{d} - 1$ because $5^{d} - 1 = 2^{a - 1}$. Thus, by evaluating mod $3$, $d$ must be odd. If $d > 1$, this is impossible as $5^{d} - 1 = (5 - 1)(5^{d - 1} + 5^{d - 2} + \dots + 5 + 1)$ and the latter factor has an odd prime factor, contradicting $5^{d} - 1$ is a power of $2$. Thus $d = 1$ and so $c = 2$, implying that $2^{a}3^{b} = 24$ so $a = 3$ and $b = 1$. Hence, the only solution is $(a, b, c, p) = (3, 1, 2, 3)$.