Solution: Note that for any $z \in \mathbb{C} $ and $m,n \in \mathbb{Z} $, \begin{align*} f(z+1) = f(z) & \implies f(z + 2) = f(z + 1) = f(z) \\ & \implies f(z + m) = f(z). \\[1ex] f(z + \sqrt{2}) = f(z) & \implies f(z + m + n \sqrt{2} ) = f(z). \end{align*} Combining these two we get \[ f(z + m + n \sqrt{2} ) = f(z), \quad z \in \mathbb{C} ,\ m,n \in \mathbb{Z} . \] Put $z = 0$ and consider the function \[ g(z) = f(z) - f(0). \] Thus, \[ g(m + n \sqrt{2} ) = f(m + n\sqrt{2} ) - f(0) = f(0 + m + n \sqrt{2} ) - f(0) = 0. \]

Now note that the set $\left\{ m + n\sqrt{2} : m, n \in \mathbb{Z} \right\} $ has many limit points, (indeed it is dense in $\mathbb{R} $). For example $\sqrt{2} $ is a limit point. To see this, take $\epsilon >0$. From the Archimedean property choose $n\in \mathbb{N} $ such that $\frac{1}{n} < \epsilon$. Now we choose $m \in \mathbb{N}$ such that \[ \left\vert \frac{m}{n} - \sqrt{2} \right\vert < \frac{1}{n^2} \implies \left\vert m - n\sqrt{2} \right\vert \leq \frac{1}{n} < \epsilon . \] Hence $g$ vanishes on a set with limit point and so \[ g(z) = 0 \quad \forall\ z\in \mathbb{C} . \] Therefore, \[ f(z) = f(0) \quad \forall\ z\in \mathbb{C} , \] and hence $f$ is a constant function.