Does $a^2+b^2=1$ have infinitely many solutions for $a,b\in\mathbb{Q}$?

Just to make explicit something which was left implicit (or hyperlinked) in the other answers: for any rational $t$ we have

$\left(\frac{1-t^2}{1+t^2}\right)^2 + \left(\frac{2t}{1+t^2}\right)^2 = 1$.

This is just what you get by projecting the $y$ axis through the point $(-1,0)$ on the circle $x^2+y^2=1$.

Generally, if a conic curve with rational coefficients has one rational point $\rm\:P\:$ then it has infinitely many, since any rational line through $\rm\:P\:$ will intersect the curve in another point, necessarily rational, since if one root of a rational quadratic is rational then so is the other. Thus by sweeping lines of varying rational slopes through $\rm\:P\:$ we obtain infinitely many rational points on the conic. Further, projecting these points onto a line leads to a rational parametrization of the conic. For a very nice exposition see Chapter 1 of Silverman and Tate: Rational Points on Elliptic Curves.

For Pythagorean Triples there is even more structure. For example one may employ ascent in the Ternary Tree of Pythagorean Triples to simply and beautifully generate them all. Picking one simple ascending path yields this formula:

$\rm\quad\ (x,\:x+1,\:z) \to (X,\:X+1,\:Z),\ \ \ X\ =\ 3\:x+2\:z+1,\ \ \ Z\ =\ 4\:x+3\:z+2\:.\ \ $ For example

$\rm\quad (3,4,5)\to (20, 21, 29)\to (119, 120, 169)\to (696, 697, 985)\to (4059, 4060, 5741)\to\cdots$

Since $\rm\ a^2 + b^2 =\: c^2\ \Rightarrow\ (a/c)^2 + (b/c)^2 =\: 1\ $ this yields infinitely many solutions to your equation.

There are infinitely many (primitive) solutions to $$a^2 + b^2 = c^2$$ with $a$, $b$, and $c$ positive integers. Given any such solution, dividing through by $c^2$ gives $$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1.$$

Since $\gcd(a,c) = \gcd(b,c)=1$, this is already expressed in least terms, so distinct primitive pythagorean triples give distinct rational solutions.