Levieee wrote:

Assume $n + a \neq n + b$ . Then we have:
$n + a \mid k \quad \text{and} \quad k \mid n + b,$ or it could also be that $k \mid n + a \quad \text{and} \quad n + b \mid k$ .

This part is certainly wrong, since $(n+a)(n+b)=k^2$ can imply $n+a \leq k$ and $n+b \geq k$ or vice-versa but not the divisibility condition above. You can find many counter examples to this.

Another finish from this part can be that let $\gcd(n+a,n+b)=d$ . (of course, $n+a \neq n+b$ )
So, $n+a = dx$ and $n+b=dy$ , $\gcd(x,y)=1$
This gives us that, $d^2xy = k^2$
Which means that $xy=l^2$ and $x=u^2$ and $y=v^2$ , $\gcd(u,v)=1$ .
This means that, $b-a=d(v^2 - u^2)$ for a particular $n$ .
Since, the original condition needs to be true for infinitely many $n$ , by infinite PHP we will have infinitely many $n$ for which we get the same value of $d$ for a particular value of $k$ .
So, for infinitely many $n$ , which in turn give infinitely many value of the pair $(u,v)$ , the value $\frac{b-a}{d}$ is constant, say $t$ .
From the above argument, since infinitely many pairs $(u,v)$ satisfy $t = v^2-u^2 = (v-u)(v+u)$ , where $v,u$ are integers, this means that $t$ has infinitely many factors $\implies$ $t=0$ , i.e., $b=a$ which contradicts that $a$ and $b$ are distinct.