, 2022

Obviously, $p$ is odd, so $p \ge 3$. If $c = 1$, then
$$p + 3 = 2^a p^b \ge 2p \ge p + 3,$$
This can only happen when $p = 3$, $a = b = 1$, giving one solution $(p, a, b, c) = (3, 1, 1, 1)$. In what follows, we assume that $c \ge 2$.

Case 1: $c$ is odd. Assume that $q$ is a prime factor of $c$. Since $(p+2)^q+1 \mid (p+2)^c+1$,
$$(p + 2)^q + 1 = 2^\alpha p^\beta. \qquad (1)$$
Obviously, $\alpha > 0$. Note that $(p+2)^q + 1 = (p+3)A$, where
$$A = (p+2)^{q-1} - (p+2)^{q-2} + \cdots + 1 > (p+2)^{q-1} - (p+2)^{q-2} = (p+2)^{q-2}(p+1) > p^{q-1},$$
and $A$ is odd, so $A$ is a power of $p$. So $A \ge p^q$ and $\beta \ge q$. Taking both sides of (1) modulo $p$ gives $2^q \equiv -1 \pmod p$, so the order of $2 \pmod p$ is 2 or $2q$.

If the order of $2 \pmod p$ is 2, then $p = 3$. In this case, $(1)$ is $5^q + 1 = 2^\alpha 3^\beta$. Since $5^q + 1 \equiv 2 \pmod 4$, we have $\alpha = 1$. By Lifting-the-exponent Lemma, $v_3(5^q + 1) = v_3(5+1) + v_3(q) \le 2$, so $\beta \le 2$. One checks that $\beta = 1, 2$ does not satisfy the equation.

If the order of $2 \pmod p$ is $2q$, then $2q \mid p-1$. So $q \le \frac{p-1}{2} < \frac{p}{2}$. Dividing both sides of (1) by $p^q$ and using $(1+\frac{1}{x})^x < e$, $x \ge 1$, we have
$$2^\alpha p^{\beta-q} = \left(1 + \frac{2}{p}\right)^q + p^{-q} < \left(1 + \frac{2}{p}\right)^{\frac{p}{2}} + p^{-q} < e + 3^{-3} < 3.$$
So $\beta = q$ and $\alpha = 1$. Using $2 \cdot p^q = (p+2)^q + 1 = (p+3)A$, we get $A = p^q$ and $p+3=2$. This is a contradiction.

Case 2: $c$ is even. Assume that $2^d \nmid c$ with $d \ge 1$. Then $(p+2)^{2d} + 1 \mid (p+2)^c + 1$. So
$$(p + 2)^{2d} + 1 = 2^\alpha p^\beta.$$
Since $(p+2)^{2d} + 1 \equiv 2 \pmod 4$, $\alpha = 1$.
$$(p + 2)^{2d} + 1 = 2 \cdot p^{\beta}. \qquad (2)$$
Taking both sides of (2) modulo $p$ gives $2^{2d} \equiv -1 \pmod p$. So the order of $2 \pmod p$ is $2^{d+1}$. Thus, $2^d < \frac{p}{2}$. Since
$$p^{\beta+1} > 2 \cdot p^{\beta} = (p+2)^{2d} + 1 > p^{2d},$$
we have $\beta \ge 2^d$. Dividing both sides of (2) by $p^{2^d}$, we have
$$2 \cdot p^{\beta-2^d} = \left(1 + \frac{2}{p}\right)^{2^d} + p^{-2^d} < \left(1 + \frac{2}{p}\right)^{\frac{p}{2}} + p^{-2^d} < e + 3^{-2} < 3,$$
so $\beta = 2^d$.

If $d \ge 2$, then $2^{d+1} \mid p-1$ implies that $p \equiv 1 \pmod 8$. Using (2), we have $p^{2^d} - 1 = (p+2)^{2^d} - p^{2^d}$. Analyzing the number of factors 2 on both sides of the equality, we deduce that
$$v_2(p^{2^d} - 1) = v_2(p^2 - 1) + d - 1 \ge d + 3.$$
Yet,
$$v_2((p+2)^{2^d} - p^{2^d}) = v_2((p+2)^2 - p^2) + d - 1 = v_2(2) + v_2(2p+2) + d - 1 = d + 2,$$
giving a contradiction. So $d=1$ and $2p^2 = (p+2)^2 + 1$, which implies that $p=5$. Going back to the original equation, we have $c = 2k$, with $k$ odd. But $a=1$, so we have
$$2 \cdot 5^b = 7^{2k} + 1.$$
Using Lifting-the-exponent Lemma again, we have $b = v_5(7^{2k} + 1) = v_5(7^2 + 1) + v_5(k) \le 2 + \frac{k}{5}$. If $k \ge 3$, then
$$2 \cdot 5^b \le 2 \cdot 5^{2+\frac{k}{5}} = (7^2 + 1) \cdot 5^{\frac{k}{5}} < 7^{2k} + 1,$$
giving a contradiction. So $k=1$, and thus $c=2$ and $b=2$. This gives another solution $(p, a, b, c) = (5, 1, 2, 2)$.

To sum up, there are two solutions of $(p, a, b, c)$: $(3, 1, 1, 1)$ and $(5, 1, 2, 2)$.