$\sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}} = \frac{1+\sqrt{5}}{2} = \phi$, is this a coincidence?

Any expression of the form $\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$ will have a valueof the form $$ \frac{-1\pm\sqrt{1+4n}}{2} $$ so it is no coincidence that you get a simple answer like you got.

As to whether it is a coincidence that this is the golden ratio, pretty much the defining expression for the golden ratio comes from a rectangle out of which a square is cut, leaving a similar rectangle -- and that diagram immediately gives you $$ \frac{1+x}{x} = \frac{x}{1} \implies 1+x = x^2$$ So although a bit of opinion creeps in, I'd say this is no coincidence.

If you play around a little more, you will also notice that: $$ \frac{1+\sqrt{5}}{2} = 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+ \ldots} } } } $$

Which simplifies to $x = 1+ \frac 1x \implies x^2=x+1$.

It's no coincidence. I mean to say, it comes directly from the equation itself.

Just to give you another example: The equation $x^2 = 4+x$ is satisfied by the fraction $\frac{1+\sqrt{17}}{2}$. Now, we can use the same logic to extend this fellow: $$ x = \sqrt{4 + x} = \sqrt{4 + \sqrt{4 + \sqrt{4 + \sqrt{4 + \ldots}}}} $$

While at the same time, this also expands as a continuous fraction, namely: $$ x = 1 + \frac{4}{x} = 1 + \frac{4}{1 + \frac{4}{1 + \frac{4}{1 + \frac{4}{1 + \ldots} } } } $$

You see, it's not a coincidence, yet it's wonderful.

The question arises: Can we do this with other quadratic polynomials?

Take for example, $ax^2+bx+c=0$. Then $ax^2 = -bx-c$ and $x^2 = -\frac{b}{a}x -\frac{c}{a}$.

This will expand now in an interesting way: $$ x = \sqrt{-\frac{c}{a}-\frac{b}{a}x} = \sqrt{-\frac{c}{a}-\frac{b}{a}\sqrt{-\frac{c}{a}-\frac{b}{a}\sqrt{-\frac{c}{a}-\frac{b}{a} \ldots}}} $$

And as a continuous fraction: $$ x = -\frac{b}{a} - \frac{c}{ax} =-\frac{b}{a} - \frac{c}{a(-\frac{b}{a} - \frac{c}{a(-\frac{b}{a} - \frac{c}{a \ldots} )}) } $$

That is your license to play around. Please do so. Also, see what you get if $ax^3+bx^2+cx+d=0$, and if you can find something interesting here do comment.

Let $a_n$ include $n$'s 1. Then $a_{n+1}=\sqrt{1+a_n}$. One can show by induction that $a_n$ is increasing and is bounded from above by $3$. Induction steps: $$ a_{n+1}=\sqrt{1+a_n}\geq\sqrt{1+a_{n-1}}=a_n; a_{n+1}=\sqrt{1+a_n}<\sqrt{1+3}=2<3. $$ So there exists $L=\lim a_n$ satisfying $L=\sqrt{1+L}$, which implies $L$ equals the value you find.