Maximizing the gain of a matrix I am interested in following problem:

$$\max_x\left(\frac{\|y\|}{\|x\|}\right) \qquad \text{subject to} \qquad y = Mx$$

Where $y$ and $x$ are vectors of same shape and $M$ is a square matrix and $||\cdot||$ is some meaningful norm (euclidean probably?). 
My gut feeling makes me suspect that the eigenvector with the maximal eigenvalue might have something to do with this, but I could not find anything particular online. Is this true, or is there any other general statement that can be made about the solution to this problem? Are there practical applications that benefit from solutions to this (I could imagine that this could come up in signal processing in some way)?
 A: In the general case, if the norms were different, one could consider to look to operator norms, as suggested by @kimchi lover.
But here, one can assume implicitly that the norm you consider in the numerator and the denominator is the quadratic norm. 
if such is the case, there is a very precise answer.
Let us square the quotient $\frac{\|Mx\|}{\|x\|}$ :
$$\frac{\|Mx\|^2}{\|x\|^2}=\frac{(Mx)^T(Mx)}{x^Tx}=\frac{x^T(M^TM)x^T}{x^Tx}$$
($T$ = transpose).
The last expression is called a Rayleigh quotient of matrix $M^TM$ https://en.wikipedia.org/wiki/Rayleigh_quotient.
Matrix $M^TM$ being symmetric semi-definite positive, let us denote its $\geq 0$ eigenvalues like this:
$$0 \leq \lambda_{1} \leq ... \leq \lambda_{n}.$$
A classical theorem says that a Rayleigh quotient takes all values between $\lambda_{1}$ and $\lambda_{n}$ inclusively.
Thus, taking the square root (reminder : we had previously squared the quotient), the result is

$$\sqrt{\lambda_{max}(M^TM)} \ \ \ \ \ \ (1)$$

(the square root of the largest eigenvalue of $M^TM$.)
If you are familiar with singular values and SVD (Singular Value Decomposition), you will realize that (1) can be described plainly as... 

The largest singular value.

