Can we determine if $y(x)=frac {c-x}{x+1}$ has any strictly positive integer solutions?

It was not perfectly clear to me whether this was a question about suitable $x$ (any solutions other than $x=0$?) or about suitable $c$ (listing various $c$ for which $x,y$ could be found). I provide an answer about $x$, and the implications for $c$.

$y$ is an integer when $c=k(x+1)+x$, viz: $y=frac{k(x+1)+x-x}{x+1}=k$.

However, since $x$ and $x+1$ have different parities, $c$ cannot be even when $x$ is odd. When $x$ is even, $k=2n$ must be even for $c=2n(x+1)+x$ to be even. In that case, $y=frac{2n(x+1)+x-x}{x+1}=2n$.

In sum, solutions exist for all $x$ and $y$ being any even integers; $c$ is limited to being of the form $c=2n(x+1)+x$ for any particular chosen $x$.

A slightly different wording than John Omielan's answer (but same idea) may make this more intuitively clearer.

$y = frac {c-x}{x+1}$ will have solutions when $x+1|c-x$.

And $x+1|c-x iff x+1|(c-x)+(x+1)= c+1$.

So for any $k|d=c+1$, we will have solutions $x = k-1$ (and $y = frac {c-(k-1)}{(k-1)+1}=frac {(c+1)-k}k = frac dk -1$)

However if $d=c+1$ is prime we either $x= 0$ or $x = d-1=c$ (in which case $y=0$).