Suppose $p$ is an odd prime. I wish to show that the number of solutions to $y^2 \equiv x^2 + 1 \pmod{p}$ is $$p + \sum_{k=0}^{p-1}\left(\frac{k^2 +1}{p}\right).$$ I know that, for any $a$, $y^2 \equiv a \pmod{p}$ admits 1 + $\left(\frac{a}{p}\right)$ solutions (mod $p$). It is thus 'intuitively' true that $y^2 \equiv x^2 + 1 \pmod{p}$ should admit $$\sum_{x=0}^{p-1}1 + \left(\frac{x^2 +1}{p}\right),$$ solutions, where we run over the distinct residue classes (mod $p$) that $x$ can assume.
Is this a sufficient argument? Or is there a way by which to formalise the proof?
What you did is fine, but this can be made more explicit.
Rewrite the equation as $(y+x)(y-x)\equiv 1\mod p$. Since $p\not=2$, the following variable change is a bijection: $$\begin{cases} u = y+x \\ v = y-x \end{cases}\Leftrightarrow \begin{cases} x = \frac{u-v}{2} \\ y = \frac{u+v}{2} \end{cases}$$
So the equation is equivalent to $uv\equiv 1 \mod p$ which has exactly $p-1$ solutions.