This answer uses Umbral calculus.

Given the formal variable $\,c,\,$ define the formal
power series linear operator

$$ L\!\!\left[\sum_{n=0}^\infty a_nc^n\right] :=
\sum_{n=0}^\infty a_nC_n. $$

Define the Catalan number generating function

$$ C(x) := L\left[\frac1{1-c\,x}\right] =
\sum_{n=0}^\infty C_n x^n =
\frac{1-\sqrt{1-4x}}{2x}. $$

Define the generating function

$$ F(x) := \sum_{n=0}^\infty
(-1)^kC_k\binom{k+2}{n-k}x^n. $$

Define the generating function

$$ f(x,y) := \sum_{n=0}^\infty \left(\sum_{k=0}^n (-1)^k{k+2\choose n-k}y^k\right)x^n. $$

Now, by previous definitions

$$ F(x) = L[f(x,c)]. $$

Verify that

$$ f(x,y) = \frac{(1 + x)^2}{ 1 + y (x + x^2) }. $$

Now get

$$ F(x) = (1+x)^2 C(-(x+x^2)) = 1+x $$

from

$$ C(-(x\!+\!x^2)) \!=\!
\frac{1\!-\!\sqrt{1\!+\!4(x\!+\!x^2)}}
{-2(x\!+\!x^2)} \!=\! \frac{1-(1\!+\!2x)}{-2(x\!+\!x^2)} \!=\! \frac1{1\!+\!x}. $$

Notice, my answer to
MSE question 4317353
also uses Umbral calculus to prove another Catalan number recursion
identity. Also, the accepted answer to this question is very similar to
mine except it does not use the Umbral calculus linear operator.