There are many different ways to look at degrees of freedom. I wanted to provide a rigorous answer that starts from a concrete definition of degrees of freedom for a statistical estimator as this may be useful/satisfying to some readers:
Definition: Given an observational model of the form $$y_i=r(x_i)+\xi_i,\ \ \ i=1,\dots,n$$ where $\xi_i=\mathcal{N}(0,\sigma^2)$ are i.i.d. noise terms and the $x_i$ are fixed. The degrees of freedom (DOF) of the estimator $\hat{y}$ is defined as $$\text{df}(\hat{y})=\frac{1}{\sigma^2}\sum_{i=1}^n\text{Cov}(\hat{y}_i,y_i)=\frac{1}{\sigma^2}\text{Tr}(\text{Cov}(\hat{y},y)),$$ or equivalently by Stein's lemma $$\text{df}(\hat{y})=\mathbb{E}(\text{div} \hat{y}).$$
Using this definition, let's analyze linear regression.
Linear Regression: Consider the model $$y_i=x_i\beta +\xi_i,$$ with $x_i\in\mathbb{R}^p$ are independent row vectors. In your case, $p=2$, and the $x_i={z_i,1}$ correspond to a point and the constant $1$, and $\beta=\left[\begin{array}{c}
m\\
b
\end{array}\right]$, that is a slope and constant term so that $x_i \beta=m z_i+b$. Then this can be rewritten as $$y=X\beta+\xi$$ where $X$ is an $n\times p$ matrix whose $i^{th}$ row is $x_i$. The least squares estimator is $\hat{\beta}^{LS}=(X^T X)^{-1}X^Ty$. Let's now based on the above definition calculate the degrees of freedom of $SST$, $SSR$, and $SSE$.
$SST:$ For this, we need to calculate $$\text{df}(y_i-\overline{y})=\frac{1}{\sigma^2}\sum_{i=1}^n\text{Cov}(y_i-\overline{y},y_i)=n-\frac{1}{\sigma^2}\sum_{i=1}^n\text{Cov}(\overline{y},y_i)=n-\frac{1}{\sigma^2}\sum_{i=1}^n \frac{\sigma^2}{n}=n-1.$$
$SSR:$ For this, we need to calculate $$\text{df}(X\hat{\beta}^{LS}-\overline{y})=\frac{1}{\sigma^2}\text{Tr}\left(\text{Cov}(X(X^TX)^{-1}X^y,y\right)-\text{df}(\overline{y})$$ $$=-1+\text{Tr}(X(X^TX)^{-1}X\text{Cov(y,y)})$$ $$=-1+\text{Tr}(X(X^TX)^{-1}X^T)$$ $$=p-1.$$ In your case $p=2$ since you will want $X$ to include the all ones vector so that there is an intercept term, and so the degrees of freedom will be $1$. However note that this will equal the number of parameters when we are doing regression with multiple parameters.
$SSE:$ $(n-1)-(p-1)=n-p$, which follows linearity of $df$.
This CANNOT be done with only the means, standard deviations, correlation, and range. The range is not needed, but one other thing is needed that was mentioned in the first paragraph but not in the question in the last paragraph: the sample size, which is $250$.
The equation of the least-squares line is:
$$
\frac{y - 478}{107.2} = 0.58\left(\frac{x-424}{81.7}\right).
$$
The sum of squares of deviations from the average $y$-value is $249\cdot 107.2^2$. The proportion of that that is explained by variability in $x$-values is $0.58^2$, so the explained sum of squares is $0.58^2\cdot249\cdot 107.2^2\approx 962597.889$. The unexplained sum of squares, which is the sum of squares of residuals, is $(1-0.58^2)\cdot249\cdot 107.2^2\approx 1898870.271$. So we have this ANOVA table:
$$
\begin{array}{c|c|c|c|c}
\text{source of variability} & \text{sum of squares} & \text{degrees of freedom} & \text{mean square} & F & p \\
\hline
\text{first test} & 962597.889 & 1 & 962597.889 & 125.719 & \approx 0 \\
\text{error} & 1898870.271 & 248 & 7656.734964
\end{array}
$$
Finding the value of $p$ is generally the only part that exceeds elementary arithmetic. But in this case the $F$ value is so extreme that it's trivial to say that for all practical purposes, $p=0$.
Best Answer
As Jonathan says you are indeed correct - well spotted. Basically if you have a 1-factor ANOVA with say q levels, then you have observations indexed by $j$ and $k$: $Y_{jk}$, $j=1,....,q,k=1,...,n_{q}$. The ANOVA model is $Y_{jk}=\mu_{j}+\epsilon_{jk}$, where $\mu_{j}$ represents the unknown true factor levels you want to estimate. This model can be written in "regression" notation by indexing your variables with just one index, say $i$: $Y_{i}=\sum\nolimits_{j=1}^{q}x_{ij}\beta_{j}+\epsilon_{i}$, $i=1,....,N$, where $x_{ij}=1$ for one of the $j's$ and zero otherwise. We assume that for your $N$ regression observations we have $n_{j}$ observations with $x_{ij}=1$, and that $\sum\nolimits_{j=1}^{q}=N$. Letting $\hat{Y_{i}}=\sum\nolimits_{j=1}^{q}x_{ij}\hat{\beta_{j}}$ we see that
$SSR=\sum\nolimits_{i=1}^{N}(\hat{Y_{i}}-\bar{Y})^{2}=\sum\nolimits_{i=1}^{N}(\sum\nolimits_{j=1}^{q}x_{ij}\hat{\beta_{j}}-\bar{Y})^{2}=\sum\nolimits_{j=1}^{q}n_{j}(\hat{\beta}_{j}-\bar{Y})^{2}$.
Now due to $x_{ij}$ being zero or one we find that $\hat{\beta}_{j}=\bar{Y}_{j}$ (I mean this to denote the average of the $Y$'s where $x_{ij}=1$), thus $\sum\nolimits_{i=1}^{N}(\hat{Y_{i}}-\bar{Y})^{2}=\sum\nolimits_{j=1}^{q}n_{j}(\bar{Y}_{j}-\bar{Y})^{2}$. In ANOVA notation we have $\bar{Y}_{j}=\bar{Y}_{j\cdot}$, and so
$SSR=\sum\nolimits_{i=1}^{N}(\hat{Y_{i}}-\bar{Y})^{2}=\sum\nolimits_{j=1}^{q}n_{j}(\bar{Y}_{j\cdot}-\bar{Y})^{2}=SSTR$.
Basically ANOVA is just a restricted form of regression, the restriction being the variables are factor variables rather than continuous ones. I find it much easier to learn about regression first, and to then think of ANOVAs in this way. This is because all the theory of regression carries over to ANOVA, but the theory about the sums of squares of ANOVAs only applies to these specific regression models, and not to a general one (where continuous and factor variables are present). If you have the time it is worth learning about regression as well as ANOVAs since the theory of ANOVAs gets you thinking from a designed experiment viewpoint (randomised controlled trials), whilst regression theory is more general since it really is about you already having your data (not from a designed experiment) and wanting to analyze it.