Denesting Ramanujan!

Squaring both sides:

$x^2+2x+1=1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\ldots}}}\\ x+2=\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\ldots}}}$

which is the original equation with $x$ replaced by $x+1$ . This means that if $x$ is a solution, then so is $x+1$ . We can see that $x=0$ is a solution by simple substitution in the original formula, hence every natural number is a solution. The answer is therefore $\boxed{\text{infinitely many}}$ .