Let $x$, $y$ $\in$ $\mathbb{R}$, find all the complex numbers $z=a+bi$ satisfying $|z+x|$ $= y$.
Because $x$ and $y$ are real numbers and $z=a+bi$ then, $|z+x|=|(a+x)+bi|=\sqrt{(x+a)^{2}+b^{2}}$.
Therefore, $\sqrt{(x+a)^{2}+b^{2}}=y$. So, $(x+a)^{2}+b^{2}=y^{2}$ $\implies$ $x^{2}+2ax+a^{2}+b^{2}=y^{2}$
The problem I now find myself with is that the only solution of $z$ I'm able to provide is in terms of itself (solving for $a$ gives me an expression with $b$ and vice versa), I believe I've done something wrong or there's something I'm missing.
I've also tried giving a geometrical interpretation of the problem in order to solve it, but I've come empty handed.
A hint or and explanation of where I'm doing something wrong would be very much appreciated.
You are absolutely on the right lines, in fact you're pretty much there. An answer in terms of $a$ or $b$ is perfectly acceptable. Solving for $b$ you obtain
$$b = \pm\sqrt{y^2 -(x+a)^2} $$
and then you can just substitute this back into your general form $z= a+bi$. So for any pair of real numbers $x$ and $y$, complex numbers of the form
$$ z = a \pm i\sqrt{y^2 -(x+a)^2 }$$
satisfy the desired property that $|z + x| = y$. In other words, given $x$ and $y$ then plugging any real number $a$ into the formula above gives you a valid complex number. Hope this helps!