The best upper bound is $1$, no matter what the values of $R_1^2$ and $R_2^2$ may be.
The following discussion explains why, in three increasingly detailed ways. The first explanation gives geometric intuition, leading to a simple example. The second one translates that into a procedure to generate specific datasets that give rise to this example. The third one generalizes this procedure to show how any mathematically possible value of $R^2$ can be achieved, given arbitrary values of $R_1^2$ and $R_2^2$.
I adopt a notation in which the independent variables are named $x_1$ and $x_2$ (rather than $x$ and $z$), so that the distinction between the independent and dependent variables remains clear.
(A comment by the alert @f coppens compels me to add that these results change when one or more of the regressions does not include a constant term, because then the relationship between $R^2$ and the correlation coefficients changes. The methods used to obtain these results continue to work. Interested readers may enjoy deriving a more general answer for that situation.)
For the simple regressions (1) and (2), the $R_i^2$ are the squares of the correlation coefficients between $x_i$ and $y$. Relationships among correlation coefficients are just angular relationships among unit vectors in disguise, because the correlation coefficient of two variables $x$ and $y$ (considered as column $n$-vectors) is the dot product of their normalized (unit-length) versions, which in turn is the cosine of the angle between them.
In these geometric terms, the question asks
How close can a vector $y$ come to the plane generated by $x_1$ and $x_2$, given the angles between $y$ and the $x_i$?
Evidently $y$ can actually be in that plane, provided you put $y$ at a given angle $\theta_1$ with $x_1$ and then place $x_2$ at a given angle $\theta_2$ with $y$. When that happens, the $R^2$ for regression (3) is $1$, demonstrating there is no meaningful upper bound on $R^2$.
Geometric thinking is no longer considered rigorous, but it leads us to a rigorous example. Start with two orthogonal unit vectors $u$ and $v$, each of which is orthogonal to a vector of ones (so that we can accommodate a constant term in all three regressions). Given $R_1^2$ and $R_2^2$, let $\rho_i^2 = R_i^2$ be choices of their square roots. To place vectors $x_1$, $y$, and $x_2$ at the required angles, set
$$\eqalign{
&x_1 &= u\\&y&=\rho_1 u + \sqrt{1-\rho_1^2} v\\ &x_2 &= (\rho_1\rho_2-\sqrt{1-\rho_1^2}\sqrt{1-\rho_2^2})u + (\rho_1\sqrt{1-\rho_2^2}\sqrt{1-\rho_1^2})v.}$$
Since $u\cdot u = v\cdot v = 1$ and $u\cdot v = 0$, you can verify that $x_2\cdot x_2 = 1$ as required,
$$y\cdot x_1 = \rho_1,$$
and
$$\eqalign{
y\cdot x_2 &= \rho_1\left(\rho_1\rho_2-\sqrt{1-\rho_1^2}\sqrt{1-\rho_2^2}\right) + \sqrt{1-\rho_1^2}\left(\rho_1\sqrt{1-\rho_2^2}\sqrt{1-\rho_1^2}\right) \\
&= \rho_2,}$$
as intended.
For a completely concrete example with $n\ge 3$ observations, start with any two $n$-vectors $u_0$ and $v_0$ which are linearly independent and linearly independent of the $n$-vector $\mathbf{1}=(1,1\ldots, 1)$. Apply the Gram-Schmidt process to the sequence $\mathbf{1}, u_0, v_0$ to produce an orthonormal basis $\mathbf{1}/\sqrt{n}, u, v$. Use the $u$ and $v$ that result. For instance, for $n=3$ you might start with $u_0 = (1,0,0)$ and $v_0=(0,1,0)$. The Gram-Schmidt orthogonalization of them yields $u = (2,-1,-1)/\sqrt{6}$ and $v=(0,1,-1)/\sqrt{2})$. Apply the preceding formulas to these for any given $R_1^2$ and $R_2^2$ you desire. This will result in a dataset consisting of the $3$-vectors $x_1$, $x_2$, and $y$ with the specified values of $R_1^2, R_2^2$, and $R^2 = 1$.
A similar approach, starting with mutually orthonormal vectors $u_0, v_0, w_0$, can be used to construct examples in which $R^2$ achieves any specified value in the interval $[\max(R_1^2, R_2^2), 1]$. Order the $x_i$ so that $R_1^2 \ge R_2^2$. Writing $y = \alpha u_0 + \beta v_0 + \gamma w_0$, $x_1 = u_0$, and $x_2 = \rho_{12}u_0 + \sqrt{1-\rho_{12}^2}v_0$, compute that $\rho_1 = \alpha$ and $\rho_2 = \alpha \rho_{12} + \beta \sqrt{1-\rho_{12}^2}$. From this, and the fact that $\alpha^2+\beta^2+\gamma^2=1$, solve and find that
$$\beta = \frac{\rho_2 - \rho_1\rho_{12}}{\sqrt{1-\rho_{12}^2}}$$
and $\gamma = \sqrt{1-\alpha^2 - \beta^2}$. For this square root to exist, $\beta$ needs to be small, but that can be guaranteed by choosing $\rho_{12}$ (the correlation between the two independent variables $x_1$ and $x_2$) to be small in size, because as $\rho_{12}$ approaches $\rho_2/\rho_1$, (which is possible because the absolute value of this ratio does not exceed $1$), $\beta$ approaches zero continuously.
The cognoscenti will recognize the relationship between the formula for $\beta$ and a certain partial correlation coefficient.
Best Answer
I think your current parameterization is sufficient to test either coefficient restriction (via the t-tests for each coefficient), but not both restrictions at the same time. To do that one can conduct an F-test between the unrestricted and the restricted model (since it is nested).
Example below in R using the
carpackage.Which then spits out a table with the RSS for each model and the degrees of freedom, needed to calculate the F-statistic.
You answered you own question in terms of testing individual coefficients, you can transform the variable on the left hand side. To replicate the
linearHypothesisfunction I trick R here by making the transformed variable:Which reproduces the earlier table. You can use this same trick to get the default summary t-test's in the regression output to test against the alternative hypotheses:
Which produces for coefficient estimates:
To test with the original values, user2864849 answers that question. You just subtract out the null from the point estimate and use the same standard error. So using the coefficients from
m1you can see belowt_estreproduces the t-statistic in model 3.