$a$ and $b$ $\in {N^*}$ : for all $n \in {N} : a^n + n | b^n +n $
We suppose that $a$ and $b$ are different . p is a prime number : $gcd(a-b,p)=1$
1) we can easily show that $ m = p(1+a) - a$ is a solution for the system $x \equiv -a \mod{p} $ and $x \equiv 1 \mod{p-1}$
2) I have to prove that $gcd(a,p)=1$
3) I have to prove that $b^m+m \equiv (b-a) \mod p$
4) I have to deduce something , I didn't do 2) and 3) but maybe I know what to deduce , Using Fermat's Theorem we can prove that $a^m+m \equiv 0 \mod{p}$ ; using the first condition we find that $ a^m + m | b^m +m $ whish means that $b^m +m$ is also congruent to $0 \mod{p}$ then we deduce that $p$ divises $b-a$ , However their gcd is $1$ , our supposition is false and $a=b$ .
The question text basically already has the answer. Here are the rest of the details for anybody who is interested.
First, note that since $a + 1 \mid b + 1$, then $b \ge a$. Next, assume that $b \neq a$, i.e., $b \gt a$. Choose any prime $p \gt b$. Since $p \gt b \gt a$, then
$$p \not{\mid} \; b \text{ and } p \not{\mid} \; a \tag{1}\label{eq1}$$
Also, since $b - a \lt p$, then
$$\gcd\left(b - a, p\right) = 1 \tag{2}\label{eq2}$$
Let
$$m = p\left(a + 1\right) - a \tag{3}\label{eq3}$$
As stated in the question, this means that
$$m \equiv -a \pmod p \tag{4}\label{eq4}$$ $$m \equiv 1 \pmod{p - 1} \tag{5}\label{eq5}$$
As such, using \eqref{eq1} and Fermat's Little Theorem, $a^{p - 1} \equiv 1 \pmod p$ giving that $a^{k\left(p - 1\right) + 1} \equiv a\left(a^{p - 1}\right)^k \equiv a \pmod p$ for any integer $k$. Also using \eqref{eq4} and \eqref{eq5} gives that
$$a^m + m \equiv a - a \equiv 0 \pmod p \tag{6}\label{eq6}$$
Therefore, this also means that $p \mid b^m + m$. Using \eqref{eq1} and Fermat's Little Theorem again, $b^{p - 1} \equiv 1 \pmod p$. Thus, using this, plus \eqref{eq4} and \eqref{eq5} again, we get that
$$b^m + m \equiv b - a \equiv 0 \pmod p \tag{7}\label{eq7}$$
However, this contradicts \eqref{eq2}, showing the initial assumption of $b \neq a$ must be false, so $b = a$.