If $f(x)^2=x+(x+1)f(x+2)$, what is $f(1)$?

Rewrite the equation to get $f(x+2)=(f(x)^2-x)/(x+1)$. Thus you may define $f$ arbitrarily in the interval $[0,2)$ and extend it to $[0,4)$ by using the equation. And so on. Choosing $f$ continuous on $[0,2)$ such that $\lim_{x\to 2}f(x)=f(0)^2$ gives (all) continuous solution. Thus the value of $f(1)$ plays no particular role.

The functional equation for $f(x)$ actually implies functional relations also for all its derivatives $$ \left\{ \matrix{ f(x)^{\,2} = x + \left( {x + 1} \right)f(x + 2) \hfill \cr 2f(x)f'(x) = 1 + f(x + 2) + \left( {x + 1} \right)f'(x + 2) \hfill \cr 2f'(x)^{\,2} + 2f(x)f''(x) = 2f'(x + 2) + \left( {x + 1} \right)f''(x + 2) \hfill \cr \quad \quad \vdots \hfill \cr} \right. $$ so that $$ \left\{ \begin{gathered} f(2) = f(0)^{\,2} \hfill \\ f'(2) = 2f(0)f'(0) - f(0)^{\,2} - 1 \hfill \\ f''(2) = 2f'(0)^{\,2} + 2f(0)f''(0) - 4f(0)f'(0) + 2f(0)^{\,2} + 2 \hfill \\ \quad \quad \vdots \hfill \\ \end{gathered} \right. $$

Therefore, being $f(x)$ continuous, we are not free to fix $f(x)\quad |\;0\le x < 2$ equal to whatever continuous function respecting only $f(2)=f(0)^2$.
Instead it shall be such as to respect the functional relation, at $x$ and $x+2$, for all the derivatives.