This nice problem was proposed by Amin Fathpour

Set $n=(kp-t)(p-1)+s$ for prime number $p$ and natural numbers $t,s$ Then we have: $n^k + a^n + c=(t+s)^k + a^s + c$ (mod p) $n^l + b^n + d=(t+s)^l + b^s + d$ (mod p) Then if $p$ divides $n^k + a^n+c$ Then we have: $(t+s)^k=-(a^s+c)$ and $(t+s)^l=-(b^s+d)$ (mod p) Now we have: $(t+s)^(kl)=(-(a^s+c))^l=(-(b^s+d))^k$ (mod p) So we have$p$ divide: $(-(a^s+c))^l-(-(b^s+d))^k$ If we increase p to a pirme number greater than $(-(a^s+c))^l-(-(b^s+d))^k$ We finde that this number is zero But how we can increase $p$? $p$ Must divide $n^k + a^n+c$ $n=(kp-t)(p-1)+s$ So p must divide $(t+s)^l + b^s + d$ And we know that Every polynomial has infintly many prime divisors! Now we find that for every natural number $s$: $(a^s+c)^l=(b^s+d)^k$ It is too easy problem! Let $j=gcd(k,l)$ set $s=kv/j$ We find that $(a^s+c)$ is a $k/j$th power But $a^s$ is $kv/j$th power If put $v$ big number we find that this equality in false! Because different between many great $k/j$th power is $c$! This is contradiction!!

can you explain me why do you choose p arbitrary ???(shurs not work with exponent)

Choose $n=p^{e} \varphi({p^e})$ to get $$c \equiv d \pmod {p^e}$$for all large enough $e$ implying $c=d$ Let $p_1>a,b$ be a prime. Then choose \begin{align*} n \equiv 0 \pmod {p_1} \\ n \equiv 0 \pmod {p_1-1} \end{align*}to get $$a \equiv b \pmod {p_1}$$or $a=b$ Similarly we can show $k=l$