The answer is ordinary least squares regression (OLS).
Because most readers will be more familiar, and comfortable with, regressing $Y$ on $X$, allow me to change notation slightly. If we switch the roles of $X$ and $Y$, take transposes, and rename $T^\prime$ as $\beta$, then in these terms the question asks:
Find $\beta$ for which the correlation between the vectors $y$ and $X\beta$ is as great as possible.
Specifically, $X$ is an $n\times k$ matrix, $y$ is an $n$-vector, and $\beta$ is a $k$-vector. Because correlation does not change when values are shifted, then without any loss of generality we may assume the average of $y$ is zero and the average of each column of $X$ likewise is zero.
The geometry of least squares immediately answers the question. By definition, the least squares solution $\hat\beta$ yields the projection $\hat y = X\hat \beta$ for which $|e|^2 = |y - \hat y|^2$ is as small as possible. Assuming $\hat y \ne 0$, we may normalize $\hat y$ to unit length (by dividing it by its own length). Let $f$ be the difference $y - \hat y / |\hat y|$. Notice that its length is the distance between $y$ and a point on the unit sphere in the space spanned by the columns of $X$. Finally, let $\theta$ be the angle between $y$ and $\hat y$. Its cosine, as is well known, is the correlation coefficient $\rho$ between any nonzero multiple of $y$ and any nonzero multiple of $\hat y$.
Here are two simple results from two-dimensional Euclidean geometry:
Since $|e| = |y|\sin\theta$ and OLS minimizes $|e|^2$, $|e|^2/|y|^2 = \sin^2\theta$ must be as small as possible. Consequently $\rho^2 = \cos^2\theta = 1-\sin^2\theta$ is as large as possible.
Since the Law of Cosines implies $|f|^2 = |y|^2 + 1 - 2|y| \cos\theta$, $|f|^2$ must be as small as possible.
These results show that the OLS solution simultaneously gives the largest possible value of $\rho^2$ and the shortest distance from $y$ to the unit sphere in the column space of $X$.
Therefore, to maximize $\rho$ itself, use either $\hat y = X\hat \beta$ or $-\hat y = X(-\hat\beta)$: exactly one of these will produce a positive value of $\rho$. Note that any positive multiple of such a solution still solves the problem of maximizing $\rho$ (but is no longer the OLS solution).
Finally, $\hat y = 0$ (which was excluded from this analysis) implies all columns of $X$ are orthogonal to $y$, whence $X\beta$ is orthogonal to $y$ for all possible values of $\beta$. In this case, the maximum value of $\rho^2$ is $0$ and all possible values of $\hat \beta$ are solutions.
Why did $y$ need to be a vector and not also a matrix $Y$? Principally because then "the" correlation between $Y$ and $X\beta$ is not defined: there are many possible correlations between vectors in the column space of $Y$ and the column space of $X$. But we could play the same game, anyway, by considerating correlations between linear combinations $X\beta$ of columns of $X$ and $Y\alpha$ of columns of $Y$. Note that the unit spheres in the $X$-space and the $Y$-space are both closed and compact. This implies the minimum distance between those spheres actually is attained, say at points $X\hat\beta$ and $Y\hat\alpha$. The same geometric analysis given above implies these two points have maximal squared correlation: that's the CCA result.